JEE MAIN - Mathematics (2021 - 26th August Morning Shift - No. 14)
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
136
Explanation
$${}^1{P_1} + 2.{}^2{P_2} + 3.{}^3{P_3} + .... + 15.{}^{15}{P_{15}}$$
= 1! + 2 . 2! + 3 . 3! + ..... 15 $$\times$$ 15!
$$ = \sum\limits_{r = 1}^{15} {(r + 1)! - (r)!} $$
= 16! $$-$$ 1
= $${}^{16}{P_{16}}$$ $$-$$ 1
$$\Rightarrow$$ q = r = 16, s = 1
$${}^{q + s}{C_{r - s}} = {}^{17}{C_{15}}$$ = 136
= 1! + 2 . 2! + 3 . 3! + ..... 15 $$\times$$ 15!
$$ = \sum\limits_{r = 1}^{15} {(r + 1)! - (r)!} $$
= 16! $$-$$ 1
= $${}^{16}{P_{16}}$$ $$-$$ 1
$$\Rightarrow$$ q = r = 16, s = 1
$${}^{q + s}{C_{r - s}} = {}^{17}{C_{15}}$$ = 136
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