If ${}^n{P_r} = {}^n{P_{r + 1}}$ and ${}^n{C_r} = {}^n{C_{r - 1}}$, then the value of r is equal to :
Solution
$${}^n{P_r} = {}^n{P_{r + 1}} \Rightarrow {{n!} \over {(n - r)!}} = {{n!} \over {(n - r - 1)!}}$$<br><br>$\Rightarrow (n - r) = 1$ .....(1)<br><br>${}^n{C_r} = {}^n{C_{r - 1}}$<br><br>$\Rightarrow {{n!} \over {r!(n - r)!}} = {{n!} \over {(r - 1)!(n - r + 1)!}}$<br><br>$\Rightarrow {1 \over {r(n - r)!}} = {1 \over {(n - r + 1)(n - r)!}}$<br><br>$\Rightarrow n - r + 1 = r$<br><br>$\Rightarrow n + 1 = 2r$ ..... (2)<br><br>From (1) and (2), $2r - 1 - r = 1 \Rightarrow r = 2$
About this question
Subject: Mathematics · Chapter: Permutations and Combinations · Topic: Fundamental Counting Principle
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