Question

You must choose four administrators from a group of 10 and four students from a group of 20. In how many ways can this happen?

Answer 1

We have to select 4 administrators from a group of 10. This comes under the chapter Combinations.
In how many ways can we select 1^st administrator: 10
In how many ways can we select 2^nd administrator: (10 -1) = 9 (one person is unavailable as he has already been selected as an administrator)
In how many ways can we select 3^rd administrator : (9 - 1) = 8
Similarly, in how many ways can we select 4^th administrator: (8 - 1) = 7
So altogether the selection can be done in 10*9*8*7 = 5040 Ways.
Now, their selection order does not matter. The 1st administrator can go in any of 4, 2nd in any of 3, 3rd in any of 2 and 4th in the only vacant position. So there are 4*3*2*1 ways which should not be considered.
Hence no of selections = 5040/(4*3*2*1) = 210 Ways

In terms of mathematics, everything we did is called as Combinations and can be written as ^₁₀C₄ which equals to 10!/(4!*(10-4)!)

In this similar way, 4 students from a group of 20 can be selected in ²⁰C₄ ways

= 20!/(4!*(20-4)!)

= 20!/(16!*4!)

= 4845 Ways

So total number of ways to select both the cases = 210*4845 = 1017450 Ways

Answer: choosing four administrators from a group of 10 and four students from a group of 20 can happen in 1017450 Ways

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