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.
[Math] In how many ways can we arrange 40 boys and 20 girls in 5 groups of 12 members each, so that each group contains at least one girl.
combinationscombinatoricspermutations
Best Answer
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 $$\frac {60!}{12!^55!}-{40 \choose 12}\frac {48!}{12!^44!}+\frac 12{40 \choose 12}{28 \choose 12}\frac{36!}{12!^33!}-\frac 1{3!}{40 \choose 12}{28 \choose 12}{16 \choose 12}\frac{24!}{12!^22!}$$