If {}^9P_5 + 5 cdot {}^9P_4 = {}^{10}P_r, the | vedclass

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(C) We know the identity ${}^nP_r + r \cdot {}^nP_{r-1} = {}^{n+1}P_r$. Alternatively, calculating the values: ${}^9P_5 + 5 \cdot {}^9P_4 = \frac{9!}{

Vedclass · Accepted Answer

$5$

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(C) We know the identity ${}^nP_r + r \cdot {}^nP_{r-1} = {}^{n+1}P_r$. Alternatively, calculating the values: ${}^9P_5 + 5 \cdot {}^9P_4 = \frac{9!}{(9-5)!} + 5 \cdot \frac{9!}{(9-4)!}$ $= \frac{9!}{4!} + 5 \cdot \frac{9!}{5!} = \frac{9!}{4!} + 5 \cdot \frac{9!}{5 \cdot 4!}$ $= \frac{9!}{4!} + \frac{9!}{4!} = 2 \cdot \frac{9!}{4!}$ $= \frac{2 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5}{1} = 30240$ Now, ${}^{10}P_r = \frac{10!}{(10-r)!}$. For $r=5$, ${}^{10}P_5 = \frac{10!}{5!} = 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 = 30240$. Thus, $r = 5$.

If ${}^9P_5 + 5 \cdot {}^9P_4 = {}^{10}P_r$, then the value of $r$ is:

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