Question

Two packs of $52$ cards are shuffled together. The number of ways, in which a man can be dealt $26$ cards, so that he does not get two cards of same suit & same denomination, is :

• A. ${}^{52}{C}_{26}\times 2^{26}$
• B. ${}^{104}{C}_{26}$
• C. $2\times {}^{52}{C}_{26}$
• D. ${}^{104}{C}_{26}\times 2^{26}$

Answer 1

Answer: (A)

First of all staple identical cards and select $26$ cards ${=}^{52}{C}_{26}$---(1)

But each of the $26$ cards can be selected in $2$ ways (belonging to either of the two packs)

These number of ways are $=2^{26}$---(2)

Hence, required number of ways of selection ${=}^{52}{C}_{26}\times 2^{26}$

(OR)

$N={}^{52}{C}_{26}\times 2\times 2\times 2\dots 2$[up to 26]

$={}^{52}{C}_{26}\times 2^{26}$
Hence, Option D is correct.