subset selections

Question

the subsets A⊂B⊂C of a universal set U, except assume that n(A)=5 and n(C)=10.

 

in how many different way can you select B if 

 

1. n(B)=6

 

2. n(B)=7

Answer 1

As A is a subset of B, B has to have 5 elements.  As B is a subset of C, which has 10 elements, then there is no choice on 5 elements of B, as they are in A.  However, we can have various possibilities for the other 10 - 5 = 5.

Thus, for n(B) = 6, we are really selecting 1 item out of 5.  This is C(5, 1) = 5

For n(B) = 7, we are selecting 2 out of 5.  This is C(5, 2) = 5*4/(2*1) = 10

 

If you want to check this result, let A have elements A, B, C, D, E and C have elements A, B, C, D, E, F, G, H, I, J

For n(B) = 6, we have the sets

{A, B, C, D, E, F}, {A, B, C, D, E, G}, {A, B, C, D, E, H}, {A, B, C, D, E, I}, {A, B, C, D, E, J}

For n(B) = 7, we have the sets

{A, B, C, D, E, F, G}, {A, B, C, D, E, F, H}, {A, B, C, D, E, F, I}, {A, B, C, D, E, F, J}, {A, B, C, D, E, G, H},

{A, B, C, D, E, G, I}, {A, B, C, D, E, G, J}, {A, B, C, D, E, H, I}, {A, B, C, D, E, H, J},  {A, B, C, D, E, I, J}

Answer 2

1. how many ways can you choose 6 elements from 10 elements ?

 10!/6!(10-6)!=10!/6!4!=10*9*8*7/4*3*2*1= (you can simplify to make the calculations easier)

10*9*8*7/4*3*2*1=10*3*7=210 ways

 

2. choose 7 from 10

  10!/7!(10-7)!=10!/7!3!=10*9*8/3*2*1=5*3*8=120 ways