Question

Let $f(n,k)$ denote the number of ways in which $k$ identical balls can be coloured with $n$ colours so that there is at least one ball of each colour. Then $f(n,2n)$ must be equal to

• A. ${}^{2n}{C}_n$
• B. ${}^{2n-1}{C}_{n+1}$
• C. ${}^{2n-1}{C}_n$
• D. none of these

Answer 1

The correct option is C ${}^{2n-1}{C}_n$

$f(n,2n)=$ colouring $2n$ identical balls with $n$ different colours which is equal to number of positive integer solutions of $x_1+x_2+...+x_n=2n$
where $x_1,x_2,\cdots ,x_n\ge 1$
Hence $(x_1-1)+(x_2-1)+...+(x_n-1)=n$
$y_1+y_2+y_3+\cdots +y_n=n$ where $y_i=x_i-1\ge 0\text{for}i=1,2,3\cdots n$
which must be equal to ${}^{n+n-1}{C}_{n-1}={}^{2n-1}{C}_n$