JEE MAIN - Mathematics (2021 - 25th July Evening Shift - No. 16)
If $${}^n{P_r} = {}^n{P_{r + 1}}$$ and $${}^n{C_r} = {}^n{C_{r - 1}}$$, then the value of r is equal to :
1
4
2
3
Explanation
$${}^n{P_r} = {}^n{P_{r + 1}} \Rightarrow {{n!} \over {(n - r)!}} = {{n!} \over {(n - r - 1)!}}$$
$$ \Rightarrow (n - r) = 1$$ .....(1)
$${}^n{C_r} = {}^n{C_{r - 1}}$$
$$ \Rightarrow {{n!} \over {r!(n - r)!}} = {{n!} \over {(r - 1)!(n - r + 1)!}}$$
$$ \Rightarrow {1 \over {r(n - r)!}} = {1 \over {(n - r + 1)(n - r)!}}$$
$$ \Rightarrow n - r + 1 = r$$
$$ \Rightarrow n + 1 = 2r$$ ..... (2)
From (1) and (2), $$ 2r - 1 - r = 1 \Rightarrow r = 2$$
$$ \Rightarrow (n - r) = 1$$ .....(1)
$${}^n{C_r} = {}^n{C_{r - 1}}$$
$$ \Rightarrow {{n!} \over {r!(n - r)!}} = {{n!} \over {(r - 1)!(n - r + 1)!}}$$
$$ \Rightarrow {1 \over {r(n - r)!}} = {1 \over {(n - r + 1)(n - r)!}}$$
$$ \Rightarrow n - r + 1 = r$$
$$ \Rightarrow n + 1 = 2r$$ ..... (2)
From (1) and (2), $$ 2r - 1 - r = 1 \Rightarrow r = 2$$
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