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There are \mathrm{p} letters \mathrm{a,q} letters \mathrm{b, r}  letters \mathrm{c}. The number of ways of selecting \mathrm{k} letters out of these if \mathrm{p} \mathrm{<k<q<r} is

Option: 1

\mathrm{\frac{1}{2}(p+1)^{2}-k}


Option: 2

\mathrm{\frac{1}{2}(p+1)(2 k-p+2)}


Option: 3

\mathrm{\frac{1}{3}(p+1)(q+1)(r+1)-k}


Option: 4

 none of these


Answers (1)

best_answer

Let \mathrm{x \, a^{\prime} s, y \, b^{\prime} s\: and \: z\, c^{\prime} s} be selected. Then number of selections is equal to the number of non-negative integral solutions of

\mathrm{x+y+z=k \quad \ldots(1)}
If we take \mathrm{x=l, 0 \leq x \leq p} then \mathrm{y+z=k-l} and its number of solutions is \mathrm{k-l+1}.

Thus, the desired number of selections

\mathrm{\sum_{l=0}^{p}(k-l+1)=(k+1)(p+1)-\frac{1}{2}(p+1)}
\mathrm{p=\frac{1}{2}(p+1)(2 k-p+2)}.

Posted by

Sanket Gandhi

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