(B) Given,${ }^{15} P_8 = A + 8 \cdot { }^{14} P_7$ $\Rightarrow \frac{15!}{7!} = A + 8 \cdot \frac{14!}{7!}$ $\Rightarrow A = \frac{15!}{7!} - 8 \cdo

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(B) Given,${ }^{15} P_8 = A + 8 \cdot { }^{14} P_7$ $\Rightarrow \frac{15!}{7!} = A + 8 \cdot \frac{14!}{7!}$ $\Rightarrow A = \frac{15!}{7!} - 8 \cdot \frac{14!}{7!}$ $\Rightarrow A = \frac{15 \cdot 14!}{7!} - \frac{8 \cdot 14!}{7!}$ $\Rightarrow A = \frac{14!}{7!} (15 - 8)$ $\Rightarrow A = \frac{14!}{7!} \cdot 7$ $\Rightarrow A = \frac{14!}{6!} = { }^{14} P_8$

Class 11 Mathematics Permutation and Combination Definition of permutation, Number of permutations with or without repetition, Conditional permutations

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${ }^{15} P_8 = A + 8 \cdot { }^{14} P_7 \Rightarrow A = $

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A
${ }^{14} P_6$
B
${ }^{14} P_8$
C
${ }^{15} P_7$
D
${ }^{16} P_9$

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Class 11

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Permutation and Combination

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Definition of permutation, Number of permutations with or without repetition, Conditional permutations

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${ }^{15} P_8 = A + 8 \cdot { }^{14} P_7 \Rightarrow A = $

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There are $9$ balls to be placed in $9$ boxes. If $5$ of the balls cannot fit into $3$ specific small boxes,find the number of ways to arrange exactly one ball in each of the boxes.

If $7 \cdot ^nP_3 = 20 \cdot ^{n+1}P_2$,then $n = \dots$

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