A pack of 52 cards is divided at random into two equals parts. Find the probability that both parts will have an equal number of black and red cards.

Answer

`({26!}^(4))/({13!}^(4)xx52!)`

Explanation

We have to find the probability that out of 26 cards drawn at random from the pack of cards, 13 will be red and 13 black. The total number of different ways to draw 26 cards from 52 is `.^(52)C_(26)`. The favorable ways are those in which there will be 13 cards drawn from 26 red cards and 13 from 26 black cards. 13 red cards may be drawn in `.^(26)C_(13)` different ways and 13 black cards also in `.^(26)C_(13)` different ways. Therefore, total number of favorable ways is equal to `.^(26)C_(13) .^(26)C_(13)`. And consequently the required probability is
`P = (.^(26)C_(13).^(26)C_(13))/(.^(52)C_(26))`
`=((((26)!)/((13)!(13)!)((26)!)/((13)!(13)!)))/((((52)!)/((26)!(26)!)))=({(26)!}^(4))/({(13)!}^(4)(52)!)`