Question

Two packs of $52$ cards are shuffled together.

The number of ways in which a man can deal $26$ cards so that he does not get two cards of the same suit and same denomination is?

• A. ${}^{52}{C}_{26}\times 2^{26}$
• B. ${}^{104}{C}_{26}$
• C. $2.{}^{52}{C}_{26}$
• D. None of these

Answer 1

The correct option is A

${}^{52}{C}_{26}\times 2^{26}$

Explanation for the correct answer.

Find the number of ways

Given: $52$ cards

$26$ cards can be chosen from $52$ cards in ${}^{52}{\mathrm{C}}_{26}$ways

And every card can be selected in $2$ ways because the card can be from the first pack or second pack of $26$ cards.

Total number of ways $={}^{52}{C}_{26}\times 2^{26}$

Hence option (A) is the correct answer.