The number of ways in which the letters A, B, | vedclass

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(D) Let the number of boxes in rows $R_1, R_2, R_3$ be $n_1=3, n_2=3, n_3=2$ respectively. Total boxes $n=8$. We need to place $5$ distinct letters in

Vedclass · Accepted Answer

$5760$

Vedclass · Answer

(D) Let the number of boxes in rows $R_1, R_2, R_3$ be $n_1=3, n_2=3, n_3=2$ respectively. Total boxes $n=8$. We need to place $5$ distinct letters in $8$ boxes such that no row is empty. Total ways to place $5$ letters in $8$ boxes is $P(8, 5) = 8 	imes 7 	imes 6 	imes 5 	imes 4 = 6720$. Let $S_1, S_2, S_3$ be the sets of ways where rows $R_1, R_2, R_3$ are empty respectively. We want to find $Total - |S_1 \cup S_2 \cup S_3|$. $|S_1|$: $R_1$ is empty,so we place $5$ letters in $8-3=5$ boxes: $P(5, 5) = 120$. $|S_2|$: $R_2$ is empty,so we place $5$ letters in $8-3=5$ boxes: $P(5, 5) = 120$. $|S_3|$: $R_3$ is empty,so we place $5$ letters in $8-2=6$ boxes: $P(6, 5) = 720$. $|S_1 \cap S_2|$: $R_1, R_2$ empty,$5$ letters in $2$ boxes: $0$ ways. $|S_1 \cap S_3|$: $R_1, R_3$ empty,$5$ letters in $3$ boxes: $0$ ways. $|S_2 \cap S_3|$: $R_2, R_3$ empty,$5$ letters in $3$ boxes: $0$ ways. $|S_1 \cap S_2 \cap S_3|$: All rows empty: $0$ ways. Using Inclusion-Exclusion Principle: $|S_1 \cup S_2 \cup S_3| = (|S_1| + |S_2| + |S_3|) - (|S_1 \cap S_2| + |S_1 \cap S_3| + |S_2 \cap S_3|) + |S_1 \cap S_2 \cap S_3| = (120 + 120 + 720) - 0 = 960$. Required ways = $6720 - 960 = 5760$.

The number of ways in which the letters $A, B, C, D, E$ can be placed in the $8$ boxes of the figure below so that no row remains empty and at most one letter can be placed in a box is:

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