If $${}^1{P_1} + 2.{}^2{P_2} + 3.{}^3{P_3} + .... + 15.{}^{15}{P_{15}} = {}^q{P_r} - s,0 \le s \le 1$$, then ${}^{q + s}{C_{r - s}}$ is equal to ______________.
Answer (integer)
136
Solution
${}^1{P_1} + 2.{}^2{P_2} + 3.{}^3{P_3} + .... + 15.{}^{15}{P_{15}}$<br><br>= 1! + 2 . 2! + 3 . 3! + ..... 15 $\times$ 15!<br><br>$= \sum\limits_{r = 1}^{15} {(r + 1)! - (r)!}$<br><br>= 16! $-$ 1<br><br>= ${}^{16}{P_{16}}$ $-$ 1<br><br>$\Rightarrow$ q = r = 16, s = 1<br><br>${}^{q + s}{C_{r - s}} = {}^{17}{C_{15}}$ = 136
About this question
Subject: Mathematics · Chapter: Permutations and Combinations · Topic: Fundamental Counting Principle
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