How many combinations with 3 groups

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We need to calculate how many unique combinations we can make. In how many ways can 4 officers, a president, vice-president, secretary and treasurer be chosen from the members of the club? b. 6P3. That is a total of 7 combinations. But how do you calculate it with some kind of formula when Dec 13, 2022 · If the sequence of the 3 matters, there are (20 x 19 x 18) = 6,840 ways to line up 3 of 20. These must come from 5 combinations of 3. Now, there are 6 (3 factorial) permutations of ABC. For example, if you have just been invited to the Oscars and you have only 2 tickets for friends and family to bring with you, and you have 10 people to choose The Letter Combination Calculator uses a fundamental formula to generate combinations: N choose K = N! / (K! (N – K)!) Where: N represents the total number of letters or elements available. Aug 17, 2021 · The binomial coefficient (n k) represents the number of combinations of n objects taken k at a time, and is read “ n choose k. For example if I pick the group of 5 first would it be the same if I choose to pick the group of 3 first? Thus, for each line up, we have 5! 5! 5! ways of arranging the students in each group. For instance, if we want to find the number of subgroups containing objects taken from objects (or the combination of objects taken at a time), it is the same as asking “how many possible groups of objects can be taken from objects?”. Feb 8, 2021 · As such, it asks for how many different groups are there, where the order within the group does not count. 1) (7. The total number of people n = 8. Mar 28, 2018 · Starting with 1 2 3 we can form combinations of size 1 2 or 3. Jul 18, 2022 · A standard deck of 52 playing cards has 4 suits with 13 cards in each suit. (r + n -1)! r! × (n - 1)! This free calculator can compute the number of possible permutations and Nov 25, 2012 · For unlabelled teams, divide by 2 2. 20 C 3. Obviously the total number of combinations would be the sum of these two cases, so 10+20=30 combinations. Basically in the above equation we are trying to find all possibilities of a combination being formed using at least two elements from the same group. We know that we have them all listed above —there are 3 choices for which letter we put first, then 2 choices for which letter comes next, which leaves only 1 choice for the May 26, 2022 · Example 3; Example 4; Combination ; Example 5; Example 6; Example 7; Example 8; Consider the following counting problems: In how many ways can three runners finish a race? In how many ways can a group of three people be chosen to work on a project? What is the difference between these two problems? I'm wondering how many possible 3 character combinations can be made using the 26 letters of the alphabet, and 0-9. In Figure 4, all the possible subgroups Unique combinations (order does matter; items repeat) combinations with 1 item 4 combinations with 2 items 4^2 = 4*4 = 16 combinations with 3 items 4^3 = 4*4*4 = 64 combinations with 4 items 4^4 = 4*4*4*4 = 256 ANSWER: 340 unique combinations (order does matter; items repeat) Unique combinations (order does not matter; items repeat): In order to find the actual number of choices we take the number of possible permutations and divide by 6 to arrive at the actual answer: 7C3 = 7P3 3! = 7! 4! ∗ 3! (7. 8 combination 2. So it Will go up until You Will form a team like this where you have all the 8 players going to one team and No Players in another written like $ 8 \choose 8$. Number of ways of selecting 4 from a group of12 is 12 C 4. So in this case, you can simply get the answer without using any formulas: xy, xz, yz, xyz x y, x z, y z, x y z. n and r are same as above. So we look at another way of 3 choose 2 basically means choosing a number of sampling elements ‘r’ from the total number of elements in the given set ‘n’. I understand how to do this if the teams are labeled: $$\frac{9!}{4!3!2!}$$ Here we select k element groups from n elements, regardless of the order, and the elements can be repeated. Correct answer: 40,320. add 5 new videos, but have 12 videos to choose from. $$\frac{15!}{3!3!3!2!2!2!}\times\frac{1}{3!3!}$$ Reasoning: Permute the $15$ people, with the first three sets of three people being the groups of three and the next three sets of two people being the sets of two; remove the duplicates within each of the six groups (at the person level). Then, choose a leader for each team and there are 103 10 3 ways of doing this. Number of ways See full answer below. Click the "Calculate" button. Their count is: Ck′ (n)=(kn+k−1 )=k!(n−1)!(n+k−1)! Explanation of the formula - the number of combinations with repetition is Mar 6, 2017 · This would be the correct answer if we had been asked "How many ways are there to choose four players for the Red Team, four for the Blue Team, and three for the Yellow Team?". Modified 6 years, Jan 5, 2016 · You can chose from the first group in 3 3 way, from the second one in 2 2 ways, from the third one in 4 4 ways. 2. There are 28 people to choose from, and we need 8. In this method. -If A is used 6 times, then the maximum usage of B, C, D, and E is 2 times each. $\endgroup$ – Umberto P. Mar 24, 2017 · We have a dataset of answers, where 10 students answer at most 20 questions. So some students do not answer all questions. To list all of the combinations of objects B, D, and F, we can use the following combinations: 1. Then $\frac{n!}{r!(n-r)!} \rightarrow \frac{10!}{3!(10-3)!}$= 120. We can label the group 1 to be group 2 and group 2 to be group 3 and so on. So ABC would be one permutation and ACB would be another, for example. However, I am not getting how to apply the formulas. A sorority with 42 members needs to choose a committee with 4 members, each with equal responsibility. For each choice you have 26 options (the letters in the alphabet). 3 P 3 = 3! (3 − 3)! = 3! 0! = 3! 1 = 3 ⋅ 2 ⋅ 1 = 6 ways. Here’s a few examples of combinations (order doesn’t matter) from permutations (order matters). Therefore, the number of ways of See full answer below. By the above theorem, the number of different combinations of friend groups is given by \[ \displaystyle C_{26,3} = \frac{P_{26,3} }{ 3! } = 15,600 \nonumber\ \] Question 841542: Combinations 3. 7c4. Dec 4, 2015 · (As in the example, the count of Group 1,2 and 3 are 4, 2 and 3 respectively. At a high school track meet the 400 meter race has 12 contestants. Dec 18, 2017 · In how many ways can $15$ different things be divided into three groups containing $7,5$ and $3$ things. 3 choose 3 in 3! 3!(3 −3)! = 1 way. As before, for unlabelled, divide by 2 2. If repeats are allowed but Answer and Explanation: 1. For example, with Combinations, a foursome group with Student 4, Student 5, Student 8 and Student 2 is the same as one with Student 5, Student 2, Student 8 and Student 4. It makes sense that there are fewer choices for a combination than a permutation, since the redundancies are being removed. I wanna see in how many ways is it possible to do so. First of all we need to know how many ways we can partition 7 into three non-zero parts, and the answer is four. Each of the three places have these options. For example, there are 6 permutations of the letters a, b, c: . To calculate the number of combinations for a given set of items, follow these steps: Enter the number of items (n) and the number of items to be taken at a time (r) in the input fields. A club has 24 members. As a general formula, if you have n n groups, indicating with |Groupk| | G r o u p k | the number of elements of the k-th group, you can The numerical answer depends on how a combination lock works. With Permutations, the order within the group makes a difference. 9 combination 4. Combination: Picking a team of 3 people from a group of 10. With three bits, there are 2 ⋅ 2 ⋅ 2 = 23 = 8 2 ⋅ 2 ⋅ 2 = 2 3 = 8 possible distinct strings. For n things choosing r combinations we can count using the formula. For example: If students are named as bob, ted, tom, bill, sally, alice, anna Apr 24, 2020 · How many combinations can a group of n people form? 2. Therefore, the number of groups each containing 3 consonants and 2 vowels is \(210. FBD 6. So in fact the correct answer is: (30 15)(15 15) =(30 15) ( 30 15) ( 15 15) = ( 30 15) Additional information: Your exercise is a specefic case of distributing n n objects into k k groups such as the i i th group contains ni n i objects Taking these two factors into account, we have three possibilities for each place: high, medium and low. DBF 4. Jun 3, 2018 · 3. Jun 23, 2023 · As we discussed in the first example, each group of three friends can be regarded as a subset of three people. Sep 20, 2020 · $3$ Players in One Team $5$ in Another mathematically=> $8 \choose 3$ and so on. May 2, 2021 · C 3 1. If repeats are not allowed, then the answer is P(40, 3) P ( 40, 3). Their count is: C k′(n)= ( kn+k −1) = k!(n−1)!(n+k−1)! Explanation of the formula - the number of combinations with repetition is equal to the number Apr 21, 2021 · Arrange all $15$ people in a line and then do two steps to remove duplicates. Getting exactly two heads (combinatorics) Exactly three heads in five flips. In how many ways can a four-member committee be formed under these restrictions: a) There are no restrictions. Oct 1, 2015 · If you allow the empty set, each of Red, Blue, and Green can be in the group (subset) or not, so there are $2^3=8$ subsets. Enter the total number of objects (n) and the number of elements taken at a time (r) 3. How many combinations can the seven colors of the rainbow be arranged into groups of three colors each? 2. nCr = n!/(r!(n-r)!) 4C3 = 4!/(3!(4-3)!) = 24/(6*1) = 4 Permutations is 24. BDF 2. Dec 11, 2020 · I have a group of 20 people and need to find how many unique groups of 3 I can create. b) At least one full professor is selected. 3 choose 2 in 3! 2!(3 −2)! = 3 ways. Number of unique 12-item groups from pool of 9 items? 1. This means that the total number of combinations of 3 letters from the set of 6 letters available to us would be 6 P\(_3\)/3! ways, since each In mathematics, disordered groups are called sets and subsets. Free online combinations calculator. Jan 16, 2014 · Some values that are worth memorizing include: $$0!=1 (\text{by defintion})\\ 1!=1 \\ 2!=2=2*1 \\ 3!=6=3*2*1 \\ 4!=24=4*3*2*1 \\ 5!=120=5*4*3*2*1$$. A chart breaks down each type's strengths, weaknesses, resistances, and vulnerabilities. Calculate the number of possible combinations; This can be calculated using the combination formula: nCr = n! / (r!(n-r)!) The number of possible combinations, nCr, is 7! / 4! * (7 - 4)! = 35. As you see, with two bits, there 2 ⋅ 2 =22 2 ⋅ 2 = 2 2 possible distinct strings. How many ways can a committee of three be chosen from a group of ten people? How many ways are there to choose a president, secretary, and treasurer. There are \displaystyle 3!=3\cdot 2\cdot 1=6 3! = 3 ⋅ 2 ⋅ 1 = 6 ways to order 3 paintings. F Nov 22, 2023 · This calculation will be counting each group of 3 people more than once. Each suit is associated with a color, either black (spades, clubs) or red (diamonds, hearts) Each suit contains 13 denominations (or values) for cards: nine numbers 2, 3, 4, …. 1) 7 C 3 = 7 P 3 3! = 7! 4! ∗ 3! In a combination in which the order is not important and there are no assigned roles the number of possibilities is defined as: May 17, 2021 · Therefore, there are 20 combinations of three items that can be chosen from a group of six. Generalizing with binomial coefficients (bit advanced) Example: Lottery probability. How many different permutations are there for the top 3 from the 12 contestants? For this problem we are looking for an ordered subset 3 contestants (r) from the 12 contestants (n). My problem is the following : how many ways are there to form a commitee of 3 3 people with a group of 10 10 men ? My answer : We choose 3 3 people; there are 10 × 9 × 8 = P(10, 3) 10 × 9 × 8 = P ( 10, 3) ways of doing that. a. 🔗. How many combinations are there? I am wondering if the ordering in which you choose the groups matters. 4. Share. Mar 10, 2024 · This combination calculator (n choose k calculator) is a tool that helps you not only determine the number of combinations in a set (often denoted as nCr), but it also shows you every single possible combination (or permutation) of your set, up to the length of 20 elements. How many different groups of 3 winners can be chosen? 2 . To solve this problem using the Combination and Permutation Calculator, do the following: Enter "4" for "Subset size". Wit 32 32 bits at our disposal, there are 232 2 32 distinct strings that can be formed. A permutation is a (possible) rearrangement of objects. N. Sep 17, 2023 · Choose 3 contestants from group of 12 contestants. 10 permutations of 5. There’s no reason for any person to be considered different from any other, based on the order chosen, so this is a combination. Their number is a combination number and is calculated as follows: C k (n) = (k n ) = k! (n − k)! n! A typical example of combinations is that we have 15 students and we have to choose three. 000? If the order of the clcments does matter. Here we select k element groups from n elements, regardless of the order, and the elements can be repeated. \ _\square\) How many ways are there to select 3 males and 2 females out of 7 males and 5 females? nPr = n!/ (n−r)! where n ≥ r ≥ 0. Differently put, the order of things is not important; only the group/combination matters now in our selection. C ( 10, 3) = 10! / ( 7! ∗ 3!) = 10 ∗ 9 ∗ 8 / ( 3 ∗ 2 ∗ 1) = 120. Combinations calculator or binomial coefficient calcator and combinations formula. 7. Permutation: Picking a President, VP and Waterboy from a group of 10. Answer and Explanation: 1. nPr = n!/(n-r)! 4P3 = 4!/(4-3)! = 24/1 = 24 10 C 3. I've seen that with just the alphabet you can create somewhere around 17k different combinations of 3 letters, but I assume adding in 0-9 must increase that number substantially. Jan 2, 2010 · In other words, a combination is just like a subgroup of a group. First, decide if this is a permutation or a combination situation. The r in this example is 4. com member to unlock this answer! Create your account. We would now like to investigate the relationship between permutation and combination problems in order to derive a formula for (n k) Let us reconsider the Counting with No Order, Example 2. There are 50 total states, and we must select 6: There are an astonishing 15 million different groups of 6 states! Hopefully this gets you started with combinations. In other words, 3 choose 2 means representing all possible configuration for 3 elements in a combination of 2 elements at a time. DFB 5. C is combination. Example: Combinatorics and probability. k is logically greater than n (otherwise, we would get ordinary combinations). How many combinations of 3 can be made with 16 letters? How many possible combinations are there when rolling 2 dice? If a person can select 3 presents from the 10 presents under a Christmas tree, how many different combinations are possible? From a group of 8 men and 7 women, a committee of 6 is chosen. Every choice is indipendent. However, if you just want to separate the 30 people into 3 groups . I believe understanding exactly how to do this would help. Nov 26, 2023 · There are (186) ( 18 6) ways to choose students for the first group, then (126) ( 12 6) for the second group (there are 18 − 6 = 12 18 − 6 = 12 students left, and (66) ( 6 6) for the last group. (If we wish to count choosing 0 Oct 17, 2023 · A few examples. Enter "30" for "Set size". First, notice that the order in which the numbers are entered on such a lock definitely matters. The Jack, Queen and King are called “face 2! × 9! = 55. Combinations with repeat. C) Set Y = {ABC, ABD, ABE, ACD, ACE, ADE} D) If person E must be in each group, then there can be only one group. Remember: This question is not asking to select 3 students out of 20 or make group of 3. So, a total of $3\times 3\times 3 = 3^3 = 27$. Each combination of 3 groups has 3! = 6 permutations, so the final answer is $\frac{34650}{6} = 5775$ Edit per the suggestion of G Cab: For nk students divided into n groups each of which has k students, the equation would be: Exercise 7. In combinations, the name of three countries is just a single group, and the sequence or order does not matter. a b c, a c b, b a c, b c a, c a b, c b a. $\begingroup$ The number of ways to make three independent choices of 1 letter out of 4 is $4 \cdot 4 \cdot 4$. In how many ways can 7 people be seated at a round Jan 29, 2024 · How Many Pokemon Type Combinations Are There? Pokemon games have 18 types, with 3 added over the years to balance the gameplay. Subtract the empty set you excluded and you have $7$. How many will there be? Combinations with repeat Jun 6, 2017 · You can make three independent choices, one for each of the three letters. Their count is: C k′(n)= ( kn+k −1) = k!(n−1)!(n+k−1)! Explanation of the formula - the number of combinations with The problem is a simple combination. You made three binary choices to define the subset. $$ If you want the letters to be unique, the calculation changes slightly. Why is my reasoning wrong ? If you have three teams at the beginning, and assign the 30 people into the groups: First, choose the members of the three teams. \) Since each group contains 5 letters, which can be arranged amongst themselves in \(5!= 120\) ways, the required number of words is \(210\times120 = 25200. Probability & combinations (2 of 2) Example: Different ways to pick officers. For more help, consider these other lessons on combinations, or this page that explains the difference between a combination and a permutation. Two of these will be from group 1 and 2, and one from one of the other groups, but it does not matter which one. There are 10 people, you want to split them into groups of 5 and 3 and 2. Feb 9, 2018 · However, there are limitations on the usage of letters: -In each group of 12, I must use A 5-6 times, B 1-3 times, C, 1-3 times, D 1-3 times, and E 1-3 times. The fourth chosen letter is one of D to H, so there are only $5$ possible combinations in this case. The top 3 will receive points for their team. Thus you can make 3 × 2 × 4 = 24 3 × 2 × 4 = 24 combinations. So sum of all these combinations is gonna be. Back to combinations, one of the most fundamental formulas for a binomial coefficient is: Sep 29, 2017 · How many combinations of groups are there possible in a set of 50 students with 10 groups of 5? Ask Question Asked 6 years, 8 months ago. For each case we count the number of ways to divide the people into these group sizes: 5,1,1 There are (75) = 21 ways to assign the people in the group of 5, after which the compositions of the remaining two groups are forced. $$ 26 \cdot 26 \cdot 26 = 26^3 = 17576. Hope it helps. C) Set Y = {ABC, ABD, ABE, ACD, ACE, ADE} For the group of 4 people, how many ways can 2 people be chosen (to make 2 possible pairs). In Combinations ABC is the same as ACB because you are combining the same letters (or people). I know that on the first part I have to use the combination formula since order doesn't matter. Number of ways of selecting 4 from remaining 8 is. Free Permutations and Combinations Calculator - Calculates the following: Number of permutation(s) of n items arranged in r ways = n P r Number of combination(s) of n items arranged in r unique ways = n C r including subsets of sets This calculator has 2 inputs. Cite. Sep 17, 2023 · Find the number of ways of choosing r unordered outcomes from n possibilities as nCr (or nCk). I choose the dividers (the lines) : $ 8-1 \choose 3-1$ $ = $ $ 7 \choose 2 $ But I was thinking, can I calculate the combination of choosing groups of 3 balls out of 8 balls to give me the answer? as followed: $ 8 \choose 3 $$ 8-3 \choose 3 $$ 8-3-3 \choose 2 $ = $ \frac {8!}{3!3!2!} $ Combinations with repeat. combinations; Share. For each bit, there are two options. how many subsets of at least 3 elements can be formed from a set of 4 elements. View this answer. Four members will be chosen at random for an interview with the school newspaper about the group. The result will be displayed on the screen, showcasing the number of Jan 18, 2024 · Similarly, this is the size of the combinations that you wish to compute. Combination Definition: A unique order or arrangement. Our Combinations Calculator is designed with simplicity and efficiency in mind. So the total number of combinations in this case is 210x3=630. Select whether you would like to calculate the number of combinations or the number of permutations using the simple drop-down menu. 3. n is the number of items. 3. K represents the number of letters or elements chosen for each combination. B) X Y. Become a Study. According to me it should be $3! \times ^{15}C_7 \times ^8C_5\times ^3C_3 =2162160 $ (note: $3!$ occurs to include the possibility of different groups (say GroupA , Group B , Group C) getting any particular combination ($7,5$ or $3$)) Oct 9, 2015 · But wait! Once you select 15 people to form group 1, there are only 15 people left to choose for group two. r is the number of items to be chosen. 6 permutation 2. â‹® â‹®. The answer, 27,405, is displayed in the "Combinations" textbox, as shown below. Conditional probability and combinations. Follow edited Dec 11, 2020 at 10:25. Combination Formula: n C r = n! r!(n - r)! where n is the number of items r is the unique arrangements. It is given that we have to choose 3 people from a group of 8 people. Hence, the answer is. The above considers the positions of the bins, so we need to remove duplicates. 4. Hence, the question is asking us to find the number of subsets of size three. P is permutations. So there should be C1030 ×C1020 C 30 10 × C 20 10 ways of doing that. How many times is each group of three being counted? From permutations, 3 chosen people can be arranged in 3 P 3 ways. We have 3! ways to label the groups. In how many ways can a group of 10 people be divided into (a) two groups consisting of 7 and 3 people (b) three groups consisting of 4,3 and 2 people? [ans: (a) 120 (b) 12600 ] Answer by Edwin McCravy(19745) (Show Source): Mar 9, 2014 · I understand the concept of combinations and permutations. Out of 171 type combinations, 9 haven't been used yet, but they may appear in future Pokemon games. Let's suppose that we have three variables: xyz(n = 3) x y z ( n = 3). If I consider all the ways 3 items can be chosen from a set of 8 items without any special conditions. Therefore, to calculate the number of combinations of 3 people (or letters) from a set of six, you need to divide 6! by 3!. However, this answer does not agree with the one provided and its compelling explanation. Order does not matter. This is 4x3/2x2=3. Explanation: Using the Fundamental Counting Principle, there would be 8 choices for the first player, 7 choices for the second player, 6 for the third, 5 for the fourth, and so on. The Mathematics Department of a small college has three full professors, seven associate professors, and four assistant professors. , 10 and Jack (J), Queen (Q), King (K), Ace (A). This calculator has 3 inputs. BFD 3. The task: we need to form 3 groups, G1, G2 and G3, and distribute the students throughout the groups. We now need to divide by 3! 3! because the order of the 3 3 groups does not matter. $ 8 \choose 0$ + $ 8 \choose 1$ + $ 8 \choose 2$ + $ 8 Oct 6, 2019 · According to the product rule we get: $${9\choose 3}{6\choose 3}{3\choose 3}=1680$$ However, the product rule considers the disjoint sets. Nov 26, 2020 · 1. Free Group Combinations Calculator - Given an original group of certain types of member, this determines how many groups/teams can be formed using a certain condition. Thus, 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 or 8! = 40, 320. So the total number of combinations is. If all of the numbers can be repeated, there are 40 ⋅ 40 ⋅ 40 40 ⋅ 40 ⋅ 40 options. -If B is used 3 times, then C can only be used 1 time, D and E could only be used 1-2 times each, and A Mar 7, 2020 · We have 1000 students and want to know what the maximum number of combinations there are of student groups where there are 3 students in each group and no student can ever be in a group with a student that they have already been in a group with. So, the number of possible committees is 28 C 8 = 3,108,105 28 C 8 = 3,108,105. Nov 8, 2021 · Case 1: All three letters A, B, C are chosen. n! r!(n − r)! So we have: 3 choose 1 in 3! 1!(3 −1)! = 3 ways. A permutation is a way to select a part of a collection, or a set of things in which the order matters and it is exactly these cases in which our permutation calculator can help you. Select three options. Select whether repeat elements are permitted. This means that every group of three has been counted 6 times in the 19656 calculation. How many diffcrent five-digit numbers can be formed with only odd numbered dieits? How manv of these numbers are sreater than 70. If the sequence of the 3 doesn't matter, there are (6,840/6) = 1,140 unique groups of 3. A) Set X has 10 possible groupings. So there are 4 4 unique combinations. This combination calculator is ideal for anyone who deals with statistics. It is asking to group 20 students in three groups. The constraint is that for each group we would like the maximum number of questions to be answered. FDB These are the six possible combinations of objects B, D, and F. Oct 5, 2019 · How many ways can you divide $9$ students into three unlabeled teams where one team contains $4$ people, one contains $3$ people and the last contains $2$ people? Unlabeled, meaning that groups with abc = bca = cba, etc. Again for the curious, the equation for combinations with replacement is provided below: n C r =. To find the number of ways to select 3 of the 4 paintings, disregarding the order of the paintings, divide the number of permutations by the number of ways to order 3 paintings. This number is (3 C 8) = 56 combinations. E) There are three ways to form a group if persons A and C must be in it. Therefore, the number of ways to arrange students in 3 equal groups is: $$\frac{15!}{5! 5! 5!}$$ But, the labeling of the groups also does not matter. Oct 7, 2016 · 1. In other words, this result shows the number of ways to create $3$ groups of $3$ people in the following scenrio: there are $3$ rooms with $9$ , $6$ and $3$ people, respectively, and $3$ from each room are So ABC would be one permutation and ACB would be another, for example. Let us learn more about permutation and combination in the below content. To calculate the number of permutations for an ordered subset of “r” elements from a set of “n” elements, the formula is: nPr = n! / (n – r)! Combinations, permutations and probabilities are important data in statistics. With the question as it is written, I think the better answer is. Oct 31, 2015 · For how many combinations, you have it. How many combinations videos can you choose? 12) How many groups of 2 sevens can be made from the 4 sevens in a deck of cards? 13) How many groups of 3 kings can be made from the 4 kings in a deck of cards? 14) How many groups of 4 hearts can be made from the 13 hearts in a deck of cards? How many ways are there to choose a committee of 8 people from this group? How many ways are to choose a committee of 8 people if the committee must consist of 2 people from each class? Answer. How many groups of four are possible? combination . 1. Case 2: Exactly two of the three letters A, B, C are chosen. “!” denotes the factorial of a number, which is the product of all Thus, 27,405 different groupings of 4 players are possible. 2. 1 2 ×(11 4) ×(7 4) ×(3 3) = 5775 1 2 × ( 11 4) × ( 7 4) × ( 3 3) = 5775. Feb 27, 2017 at 15:37 `C_5^5 times C_3^10` `= frac{5!}{5! times 0!} times frac{10!}{3! times 7!}` `= 120` So the number of committees with at least `2` prefects is: `2100 + 2520 + 1050 + 120 = 5790` Alternative Solution: The problem with the method used above is that if we have many (say `20`) to count, it would become very tedious. So, if all players are candidates to be good players, the solution should be (10 4)(6 3) = 4200 ( 10 4) ( 6 3) = 4200? @Helen: If all players are good players, and we are dividing into labelled teams, your expression is right. wy cs ma ba wu ls dg vh ue sj