Question

Determine the number of 5 card combinations out of a deck of 52 cards, if there is exactly one ace in each combination.

Answer 1

Hint: We have to select 5 cards from the deck of 52 cards in such a way that there is exactly one ace in 5 cards. Now, we know that there are 4 aces in a deck of 52 cards and exactly one of the 5 cards contain an ace so we have to select 1 card from the 4 aces. Now, we have selected one card and we have to select 4 cards. As the remaining 4 cards should not contain any ace so out of 52 cards we are remaining with 48 cards that we can select our remaining cards so select 4 cards from 48 cards and then multiply the selection of 1 ace and the remaining 4 cards.

Complete step-by-step answer:
We have given a deck of 52 cards out of which we have to find the combinations of 5 cards in such a way that exactly one card is an ace and the remaining 4 cards can be anything but not the ace.
So, we know that a deck of 52 cards contains 4 aces. Now, we are going to select a card from the 4 aces in the following combinatorial manner:
${}^4{C}_1$
In the above, we have selected 1 card out of 5 cards. Now, we have to select 4 cards from the remaining cards in such a way that these 4 cards do not contain any ace. Excluding the aces from 52 cards we are left with 48 cards so we have to select 4 cards from 48 cards in combinatorial way:
${}^{48}{C}_4$
Now, to get the number of combinations of 5 cards we have to multiply ${}^4{C}_1$ by ${}^{48}{C}_4$ which will look like as follows:
${}^4{C}_1\left({}^{48}{C}_4\right)$
We know that,
${}^n{C}_r={\displaystyle \frac{n!}{r!(n-r)!}}$
Using the above formula, we are going to solve the combinatorial expression:
$\begin{array}{rl}& {}^4{C}_1\left({}^{48}{C}_4\right)\\ & ={\displaystyle \frac{4!}{1!3!}}\left({\displaystyle \frac{48!}{4!44!}}\right)\end{array}$
The factorial of a number say n is equal to:
$n!=n(n-1)(n-2).......2.1!$
Using this factorial notation we can write the above expression as:
$\begin{array}{rl}& {\displaystyle \frac{4.3!}{1!3!}}\left({\displaystyle \frac{48.47.46.45.44!}{4!44!}}\right)\\ & =4\left({\displaystyle \frac{48.47.46.45}{4.3.2.1}}\right)\end{array}$
4 will be cancelled out from the numerator and denominator.
$\begin{array}{rl}& \left({\displaystyle \frac{48.47.46.45}{3.2.1}}\right)\\ & ={\displaystyle \frac{4669920}{6}}\\ & =778320\end{array}$
Hence, the number of 5 card combinations out of a deck of 52 cards, if there is exactly one ace in each combination is 778320.

Note: You might think why we have multiplied the selection of an ace card with non ace cards. We can add them also but why we have multiplied them. The answer is both the selection of ace cards or non ace cards are independent to each other. And we know that if the two events are independent to each other then we multiply their occurrences.