Question

A standard deck consists of \[52\] cards. How many \[5 - card\] hands are possible if the cards are all hearts?

Answer 1

Hint: This question is from the topic of permutations and combinations. In this question, first we will find out how many cards are of hearts in a deck of cards. After that using the formula of combination i.e., \[{}^n{C_r} = \dfrac{{n!}}{{r!{\text{ }} \times \left( {n - r} \right)!}}\] we will find out how many \[5 - card\] combinations are possible if the cards are all hearts. Then we will solve the expression and find the required result.

Complete step by step answer:
In this question, we are given a deck of \[52\] cards and we have to find how many \[5 - card\] hands are possible if the cards are all hearts. Now, first of all we will find out how many cards are of hearts in a deck of cards. As we know, a standard pack of cards consists of \[4\] suits and each set contains the same number of cards. So, the number of cards in each suit will be \[\dfrac{{52}}{4} = 13\]. Hence, the number of hearts in a deck of \[52\] cards is \[13\]

Now, we know that number of ways we can combine \[n\] objects taken \[r\] at a time is given by the formula:
\[{}^n{C_r} = \dfrac{{n!}}{{r!{\text{ }} \times \left( {n - r} \right)!}}\]
According to this question,
\[n = 13\] and \[r = 5\]
Therefore, we get
\[{}^{13}{C_5} = \dfrac{{13!}}{{5!{\text{ }} \times \left( {13 - 5} \right)!}}\]
\[ \Rightarrow {}^{13}{C_5} = \dfrac{{13!}}{{5!{\text{ }} \times 8!}}\]
Now, we know that
\[n! = n \times \left( {n - 1} \right)!\]
So, \[13! = 13 \times \left( {12} \right)!\]

Therefore, we have
\[ \Rightarrow {}^{13}{C_5} = \dfrac{{13 \times 12!}}{{5!{\text{ }} \times 8!}}\]
In the same way, we can write \[12!\]
\[ \Rightarrow {}^{13}{C_5} = \dfrac{{13 \times 12 \times 11!}}{{5!{\text{ }} \times 8!}}\]
Proceeding in the same way, we get
\[ \Rightarrow {}^{13}{C_5} = \dfrac{{13 \times 12 \times 11 \times 10 \times 9 \times 8!}}{{5!{\text{ }} \times 8!}}\]
On cancelling \[8!\] from both numerator and denominator, we get
\[ \Rightarrow {}^{13}{C_5} = \dfrac{{13 \times 12 \times 11 \times 10 \times 9}}{{5!{\text{ }}}}\]

Now we know that
\[n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times .... \times 3 \times 2 \times 1\]
So, \[5! = 5 \times 4 \times 3 \times 2 \times 1\]
Therefore, we get
\[ \Rightarrow {}^{13}{C_5} = \dfrac{{13 \times 12 \times 11 \times 10 \times 9}}{{5 \times 4 \times 3 \times 2 \times 1{\text{ }}}}\]
After simplifying it, we get
\[ \Rightarrow {}^{13}{C_5} = 13 \times 11 \times 9\]
\[ \therefore {}^{13}{C_5} = 1287\]

Hence, the number of hands possible consisting of only hearts is \[1287\].

Note: To solve these types of problems, students should be aware of permutation and combination concepts. Also, they should have brief knowledge about the pack of cards. And be aware of calculation mistakes because any mistake may change the value of the result. And also note that in a hand of cards problem, the order in which the cards are dealt is not relevant. Hence, a single hand is considered as one combination.