Permutations problem

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Permutations problem

 A class of 27 students needs to be divided into 9 teams of three students each? How many ways are there to do that? Thank you.
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Re: Permutations problem

 This post was updated on . 27!/((9!)^3!*3!)
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Re: Permutations problem

 hey shefali.. can you pls elaborate.. the steps behind your ans.
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Re: Permutations problem

 In reply to this post by rongmon It will be (27!)/(3!)^9 i guess...its similar to the form of arrangements of 27 persons in to 9 groups of 3 each..e.g n things of which p, q and r are of different types can be arranged in (n!/p!*q!*r!) ways...just like that in this case the problem is arrangement of 27 students in to 9 types with three in each group...!!!  "I don't ride side-saddle. I'm as straight as a submarine"
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Re: Permutations problem

 I go with Subhayu's answer. My logic is that it should be 27C3*24C3*21C3*18C3*15C3*12C3*9C3*6C3*3C3. It is easy to simplify it coz when you expand the C formula, a lot of terms cancel out and you get the same answer, ie, 27!/3!^9. Could you tell the right answer plz? I dont understand the logic behind shefali's answer.
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Re: Permutations problem

 In reply to this post by Granpa Simpson That is what i thought too. Answer is stated as 27!/9!*(3!)^9 though.
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Re: Permutations problem

 I can't figure why the 9! term in the denominator.
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Re: Permutations problem

 here we are dividing by 9! because all 9 groups are having same no. of persons.
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Re: Permutations problem

 There is a definite formula for such a question when division into groups is asked. That is (m+n)!/m!*n! Where m and n refer to the amount of each group. So here as the total is 27 and u need to divide into groups of 3. So (m+n) is 27 and m!=n!=3! . So in a way acc to the formula u get 27!/(3!*3!*3!*3!*3!*3!*3!*3!*3!) Which is same as 27!/(3!)^9  ok. vandita On 28 Apr 2014 21:02, "ViV [via Discussion forum]" <[hidden email]> wrote: here we are dividing by 9! because all 9 groups are having same no. of persons. If you reply to this email, your message will be added to the discussion below: http://discussion-forum.2150183.n2.nabble.com/Permutations-problem-tp7587982p7588001.html To start a new topic under General Discussions, email [hidden email] To unsubscribe from Discussion forum, click here. NAML
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Re: Permutations problem

 Administrator In reply to this post by rongmon Answer is indeed 27!/[(3!^9)9!] Let the set of students be {1, 2, 3, 4, ...., 27} Consider the following division into 9 teams {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}, {10, 11, 12}, {13, 14, 15}, {16, 17, 18}, {19, 20, 21}, {22, 23, 24}, {25, 26, 27}} And compare it with {{4, 5, 6}, {1, 2, 3}, {7, 8, 9}, {10, 11, 12}, {13, 14, 15}, {16, 17, 18}, {19, 20, 21}, {22, 23, 24}, {25, 26, 27}} Note that above two are giving us same division of 27 people into 9 teams but they count as two different arrangements in 27!. Likewise there are 9! arrangements, all resulting in the same division. So, the number 27! must be divided by 9!. Also, Consider the following division into 9 teams again {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}, {10, 11, 12}, {13, 14, 15}, {16, 17, 18}, {19, 20, 21}, {22, 23, 24}, {25, 26, 27}} And compare it with {{2, 1, 3}, {4, 5, 6}, {7, 8, 9}, {10, 11, 12}, {13, 14, 15}, {16, 17, 18}, {19, 20, 21}, {22, 23, 24}, {25, 26, 27}} Again the above two are two different arrangements in 27! but result in same team formations. Likewise, there are 3!^9 such arrangements that result in the same team formations. So, the number must also be divided by 3!^9. Thus, the correct answer is 27!/9!*(3!^9)
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Re: Permutations problem

 Thank you so much, Sir! In nutshell, we should never forget that arrangement of teams doesn't matter, here! So, divide by 9!
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Re: Permutations problem

 In reply to this post by Amit Goyal Thank you.