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(B) Total number of ways to place $4$ distinct balls into $4$ distinct boxes is $4^4 = 256$.For exactly one box to be empty,we must have $3$ boxes occ

Vedclass · Accepted Answer

$\frac{9}{16}$

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(B) Total number of ways to place $4$ distinct balls into $4$ distinct boxes is $4^4 = 256$.For exactly one box to be empty,we must have $3$ boxes occupied by $4$ balls. This implies one box contains $2$ balls and the other two boxes contain $1$ ball each.The number of ways to choose the empty box is $\binom{4}{1} = 4$.The number of ways to choose the box with $2$ balls from the remaining $3$ boxes is $\binom{3}{1} = 3$.The number of ways to distribute $4$ balls into these $3$ boxes such that one has $2$ balls and two have $1$ ball each is given by $\frac{4!}{2!1!1!} = \frac{24}{2} = 12$.Total favorable ways = $4 	imes 3 	imes 12 = 144$.Probability = $\frac{144}{256} = \frac{9}{16}$.

column-$I$	column-$II$
$(A)$ $A$ line from the origin meets the lines $\frac{x-2}{1}=\frac{y-1}{-2}=\frac{z+1}{1}$ and $\frac{x-\frac{8}{3}}{2}=\frac{y+3}{-1}=\frac{z-1}{1}$ at $P$ and $Q$ respectively. If length $PQ=d$,then $d^2$ is	$(p)$ $-4$
$(B)$ The values of $x$ satisfying $\tan ^{-1}(x+3)-\tan ^{-1}(x-3)=\sin ^{-1}\left(\frac{3}{5}\right)$ are	$(q)$ $0$
$(C)$ Non-zero vectors $\vec{a}, \vec{b}$ and $\vec{c}$ satisfy $\vec{a} \cdot \vec{b}=0$,$(\vec{b}-\vec{a}) \cdot(\vec{b}+\vec{c})=0$ and $2\|\vec{b}+\vec{c}\|=\|\vec{b}-\vec{a}\|$. If $\vec{a}=\mu \vec{b}+4 \vec{c}$,then the possible values of $\mu$ are	$(r)$ $4$
$(D)$ Let $f$ be the function on $[-\pi, \pi]$ given by $f(0)=9$ and $f(x)=\frac{\sin \left(\frac{9 x}{2}\right)}{\sin \left(\frac{x}{2}\right)}$ for $x \neq 0$. The value of $\frac{2}{\pi} \int_{-\pi}^\pi f(x) dx$ is	$(s)$ $5$
	$(t)$ $6$

Suppose four balls labelled $1, 2, 3, 4$ are randomly placed in boxes $B_1, B_2, B_3, B_4$. The probability that exactly one box is empty is

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