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My approach

There are 5 groups with 12 members each,so if there was condition like there should be 3 girls and 2 boys i would do

(20C3)*(40C2)

But here it is given as atleast one girl,how to solve this.

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    $\begingroup$ This looks like it calls for the inclusion-exclusion principle, subtracting arrangements with one or more boys-only groups in them. $\endgroup$ – Henning Makholm Feb 19 '14 at 18:35
  • $\begingroup$ First of all, we should find the partitions of 20 with 5 nonzero-numbers $\endgroup$ – Peter Feb 19 '14 at 18:35
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    $\begingroup$ You could use Inclusion/Exclusion. But then you need to first attack the problem of division with no restrictions. $\endgroup$ – André Nicolas Feb 19 '14 at 18:36
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    $\begingroup$ The problem needs to be specified better. Do the identities of the students matter? If I take a solution and swap two of the groups, is that the same solution or a distinct one? Does the order of students within the group matter? $\endgroup$ – NovaDenizen Feb 19 '14 at 22:47
  • $\begingroup$ @NovaDenizen This is all that is given in the question paper and this is not a research level question.Just an Examination question.So see this from that POV $\endgroup$ – techno Feb 20 '14 at 4:16
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To get the total number of ways to split the group into five groups of twelve, there are $\frac {60!}{12!^55!}\approx 2.75E36$. To get a group of twelve with all boys, there are ${40 \choose 12}\frac {48!}{12!^44!}\approx 5.49E34$ ways. As this is small, the error in double counting is smaller yet. We have double counted the cases with two groups of all boys, which is $\frac 12{40 \choose 12}{28 \choose 12}\frac{36!}{12!^33!}\approx 4.79E31$, so have to subtract that off. Now the cases with three groups of all boys have been subtracted three times and added three times, so we need to subtract them once again. That is $\frac 1{3!}{40 \choose 12}{28 \choose 12}{16 \choose 12}\frac{24!}{12!^22!}\approx 6.97E25$. The final answer is $$\frac {60!}{12!^55!}-{40 \choose 12}\frac {48!}{12!^44!}+\frac 12{40 \choose 12}{28 \choose 12}\frac{36!}{12!^33!}-\frac 1{3!}{40 \choose 12}{28 \choose 12}{16 \choose 12}\frac{24!}{12!^22!}$$

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  • $\begingroup$ Im a noob,im using Permutations and Combinations.From a layman's POV can you help me solve this somewhat like this math.stackexchange.com/questions/683053/… and this is all the data i have about the question no extra details are specified $\endgroup$ – techno Feb 20 '14 at 4:19
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    $\begingroup$ For example, the second batch, requiring a group with twelve boys, you first choose the twelve boys out of forty, then line up the remaining 48 people. There are 48! ways to line up the 48 people, but you can permute each group of 12 in 12! equivalent ways and permute the 4 groups in 4! equivalent ways. The other work the same. $\endgroup$ – Ross Millikan Feb 20 '14 at 4:23

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