There are $n$ different coupons, each of which can occupy $N (N > n)$ different envelopes, with the same probability $1 / N$
$P_{1}$ : The probability that there will be one gift coupon in each of $n$ definite envelopes out of $N$ given envelopes
$P_{2}$ : The probability that there will be one gift coupon in each of $n$ arbitrary envelopes out of $N$ given envelopes
Consider the following statements
i) $P_{1} = P_{2}$
ii) $P_{1} = \frac{n!}{N^{n}}$
iii) $P_{2} = \frac{N!}{N^{n} (N - n)!}$
iv) $P_{2} = \frac{n!}{N^{n} (N - n)!}$
v) $P_{1} = \frac{N!}{N^{n}}$
Then which of the following is true

Only i

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ii and iii

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ii and iv

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iii and v

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Solution

The correct option is B ii and iii
$P_{1} = \frac{n!}{N^{2}}$ (one gift coupons in a different envelopes)
$N \to n$ $⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} n & c o u p o n s g o i n g & i n t o N & d i f f e r e n t e n v e l o p e s \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$
$12 3... n$
$N N N N$
$= N^{n}$
$P_{2} = \frac{^{N} C_{n} n!}{N^{n}} = \frac{N!}{n! (N - n)!} \times \frac{n!}{N^{n}}$
$= \frac{N!}{N^{n} (N - n)!}$ (one gift common in each in envelopes arbitrary)
$∴ (B)$ (ii) & (iii)

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The correct option is B ii and iiiP1=n!N2 (one gift coupons in a different envelopes) N→n ⎡⎢ ⎢ ⎢ ⎢⎣ncouponsgoingintoNdifferentenvelopes⎤⎥ ⎥ ⎥ ⎥⎦123...nNNNN=Nn P2=NCnn!Nn=N!n!(N−n)!×n!Nn=N!Nn(N−n)! (one gift common in each in envelopes arbitrary)∴(B) (ii) & (iii)