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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
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
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Solution
The correct option is C 2n−1Cn
f(n,2n)= colouring 2n identical balls with n different colours which is equal to number of positive integer solutions of x1+x2+...+xn=2n
where x1,x2,⋯,xn≥1
Hence (x1−1)+(x2−1)+...+(xn−1)=n
y1+y2+y3+⋯+yn=n where yi=xi−1≥0
which must be equal to n+n−1Cn−1=2n−1Cn.
f(n,2n)= colouring 2n identical balls with n different colours which is equal to number of positive integer solutions of x1+x2+...+xn=2n
where x1,x2,⋯,xn≥1
Hence (x1−1)+(x2−1)+...+(xn−1)=n
y1+y2+y3+⋯+yn=n where yi=xi−1≥0
which must be equal to n+n−1Cn−1=2n−1Cn.
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