If { }^{1} P_{1}+2 cdot{ }^{2} P_{2}+3 cdot{ | vedclass

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(A) The given expression is $\sum_{n=1}^{15} n \cdot {}^{n}P_{n}$.Since ${}^{n}P_{n} = n!$,the expression becomes $\sum_{n=1}^{15} n \cdot n!$.We know

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$136$

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(A) The given expression is $\sum_{n=1}^{15} n \cdot {}^{n}P_{n}$.Since ${}^{n}P_{n} = n!$,the expression becomes $\sum_{n=1}^{15} n \cdot n!$.We know that $n \cdot n! = (n+1-1) \cdot n! = (n+1)! - n!$.Thus,the sum is $\sum_{n=1}^{15} ((n+1)! - n!) = (2!-1!) + (3!-2!) + \ldots + (16!-15!) = 16! - 1! = 16! - 1$.Given the expression is ${}^{q}P_{r} - s$,we have ${}^{16}P_{16} - 1 = {}^{q}P_{r} - s$.Comparing,we get $q = 16$,$r = 16$,and $s = 1$.We need to find ${}^{q+s}C_{r-s} = {}^{16+1}C_{16-1} = {}^{17}C_{15}$.${}^{17}C_{15} = {}^{17}C_{2} = \frac{17 	imes 16}{2 	imes 1} = 17 	imes 8 = 136$.

If ${ }^{1} P_{1}+2 \cdot{ }^{2} P_{2}+3 \cdot{ }^{3} P_{3}+\ldots+15 \cdot{ }^{15} P_{15}={ }^{q} P_{r}-s$,where $0 \leq s \leq 1$,then ${ }^{q+s} C_{r-s}$ is equal to .... .

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