Three boys and two girls stand in a queue. Th | vedclass

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Question

(A) Let $B$ represent a boy and $G$ represent a girl. There are $3$ boys and $2$ girls. The total number of ways to arrange $5$ people is $5! = 120$.

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) Let $B$ represent a boy and $G$ represent a girl. There are $3$ boys and $2$ girls. The total number of ways to arrange $5$ people is $5! = 120$. Let $b_i$ be the number of boys ahead of the $i$-th girl and $g_i$ be the number of girls ahead of the $i$-th girl. The condition is $b_i \ge g_i + 1$ for both girls. Let the positions of the girls be $x_1$ and $x_2$ where $1 \le x_1 For the first girl at $x_1$,the number of boys ahead is $x_1 - 1$ and girls ahead is $0$. So,$x_1 - 1 \ge 0 + 1 \implies x_1 \ge 2$. For the second girl at $x_2$,the number of boys ahead is $x_2 - 2$ and girls ahead is $1$. So,$x_2 - 2 \ge 1 + 1 \implies x_2 \ge 4$. Possible positions $(x_1, x_2)$ are $(2, 4), (2, 5), (3, 4), (3, 5), (4, 5)$. Checking the condition $b_i \ge g_i + 1$: $1$. $(2, 4): b_1=1, g_1=0 (1 \ge 1); b_2=2, g_2=1 (2 \ge 2)$. Valid. $2$. $(2, 5): b_1=1, g_1=0 (1 \ge 1); b_2=3, g_2=1 (3 \ge 2)$. Valid. $3$. $(3, 4): b_1=2, g_1=0 (2 \ge 1); b_2=2, g_2=1 (2 \ge 2)$. Valid. $4$. $(3, 5): b_1=2, g_1=0 (2 \ge 1); b_2=3, g_2=1 (3 \ge 2)$. Valid. $5$. $(4, 5): b_1=3, g_1=0 (3 \ge 1); b_2=3, g_2=1 (3 \ge 2)$. Valid. Total valid arrangements = $5 	imes (3! 	imes 2!) = 5 	imes 12 = 60$. Probability = $\frac{60}{120} = \frac{1}{2}$.

Three boys and two girls stand in a queue. The probability that the number of boys ahead of every girl is at least one more than the number of girls ahead of her is:

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