Question

We are given M urns, numbered 1 to M and n balls (n < M) and P(A) denote the probability that each of the urns numbered 1 to n, will contain exactly one ball.

$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ \left(a\right)& \text{If the balls are different and any number of balls can go to any urn then P(A)}=\underset{–}{}& \left(p\right)& \frac{1}{{}^M{C}_n}\\ \left(b\right)& \text{If the balls are identical and any number of balls can go to any urn then P(A)}=\underset{–}{}& \left(q\right)& \frac{1}{{}^{(M+n-1)}{C}_{M-1}}\\ \left(c\right)& \text{If the balls are identical but at most one ball can be put in any box, then P(A)}=\underset{–}{}& \left(r\right)& \frac{n!}{{}^M{C}_n}\\ \left(d\right)& \text{If the balls are different and at most one ball can be put in any box, then P(A)}=\underset{–}{}& \left(s\right)& \frac{n!}{M^n}\end{array}$

• A. $(a\to s,b\to q,c\to p,d\to p)$
• B. $(a\to s,b\to q,c\to r,d\to p)$
• C. $(a\to s,b\to r,c\to q,d\to p)$
• D. $(a\to s,b\to s,c\to p,d\to r)$

Answer 1

The correct option is A

$(a\to s,b\to q,c\to p,d\to p)$

$\left(a\right)n\left(s\right)=M^n;n\left(A\right)=n!\Rightarrow P\left(A\right)=\frac{n!}{M^n}$

Since any number of different balls can go in any urn, each ball can be placed in $M$ ways. Therefore sample space $=M^n$ ways. Favourable outcome would be permuatuting $n$ balls in the first $n$ urns. Therefore favourable outcome $=n!$ ways.

$\left(b\right)n\left(s\right)={{}^{M+n-1}C}_{M-1};n\left(A\right)=1\Rightarrow P\left(A\right)=\frac{1}{{}^{M+n-1}{C}_{M-1}}$

Since balls are all identical, just placing the balls would result as our sample space. We have $n$ balls and to distribute them in $M$ urns, we need to segegrate them by $M-1$ divisions. Therefore sample space $={{}^{M+n-1}C}_{M-1}$ ways. Favourable outcome would be permuatuting $n$ identical balls in the first $n$ urns. Therefore favourable outcome $=1$ ways.

$\left(c\right)n\left(s\right)={{}^{M}C}_n;n\left(A\right)=1\Rightarrow P\left(A\right)=\frac{1}{{}^M{C}_n}$

Selecting $n$ urns out of $M$ urns in ${{}^{M}C}_n$ ways. Favourable outcome would be permuatuting $n$ identical balls in the first $n$ urns. Therefore favourable outcome $=1$ ways.

$\left(d\right)n\left(s\right)={{}^{M}C}_n\cdot n!;n\left(A\right)=n!\Rightarrow P\left(A\right)=\frac{1}{{{}^{M}C}_n}$

Selecting $n$ urns out of $M$ urns and permuting the $n$ balls in them in ${{}^{M}C}_n\cdot n!$ ways. Favourable outcome would be permuatuting $n$ balls in the first $n$ urns. Therefore favourable outcome $=n!$ ways.