Question

In Example 28, the number of ways of choosing A
and B such that B contains just one element more then A, is
A. `""^(2n)C_(n-1)`
B. `3^(n)`
C. `(2^(n))^(2)`
D. `""^(2n)C_(n)`

Answer 1

Correct Answer - a
if A contains r elements, the B must contain
`(r +1)` elements. In this case, the number of ways of choosing A
and B is` ""^(n)C_(r)xx""^(n)C_(r+1) = C_(r). C_(r +1).` But , r can very from 0 to (n-1) .
So , the number of ways of choosing A and B
`sum_(r=0)^(n-1)C_(r) C_(r+1)= C_(0) C_(1) +C_(1)C_(2) +...+C_(n-1)C_(n) = ""^(2n)C_(n-1)` .