Question

In how many ways can a pack of 52 cards be divided into 4 sets, three of them having 17 cards each and the fourth just one card?
A.$\dfrac{{52!}}{{{{\left( {17!} \right)}^3} \cdot 3!}}$
B.$\dfrac{{52!}}{{{{\left( {17!} \right)}^3}}}$
C.$\dfrac{{52!}}{{51! \cdot {{\left( {17!} \right)}^3}}}$
D.$\dfrac{{52!}}{{51!}}$

Answer 1

Here, we will use the concept of combinations to solve this question. We will first find out the number of ways of selecting cards for the first set from the given 52 cards. Then we will find the number of ways of selecting cards for the second set from the remaining cards and follow the same procedure for the third and fourth set. Then we will multiply all the number of ways and divide it by $3!$ to get the required number of ways.

Formula Used:
${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, where $n$ is the total number of elements and $r$ is the number of elements to be selected.

Complete step-by-step answer:
Total number of cards in a pack $= 52$
We have to divide these 52 cards into 4 sets such that three sets have 17 cards each and the fourth set has just one card.
Now, since, we are required to do selection in this question, hence, we will use the formula of combinations.
First of all, we are required to select 17 cars for the first set out of the total 52 cards.
Hence, taking $n = 52$ and $r = 17$, the number of possible ways of selection is: ${}^{52}{C_{17}}$
Now, since, the 17 cards are already selected for the first set.
Hence, the number of cards which are remaining $= \left( {52 - 17} \right) = 35$
Therefore, out of these 35 cards we are required to select 17 cards for the second set.
Now taking $n = 35$ and $r = 17$, the number of possible ways of selection is: ${}^{35}{C_{17}}$
As more 17 cards are already selected for the second set as well, so
The number of cards which are now remaining$= \left( {35 - 17} \right) = 18$
Therefore, out of these 18 cards we are required to select 17 cards for the third set.
Hence, substituting $n = 18$ and $r = 17$, the number of possible ways of selection is: ${}^{18}{C_{17}}$
Finally, after selection of these cards, the remaining cards which are left for the selection of the fourth set $= \left( {18 - 17} \right) = 1$
Hence, we are required to select 1 card from the remaining 1 card for the last set.
Hence, substituting $n = 1$ and $r = 1$, the number of possible ways of selection is: ${}^1{C_1}$
Now we will find the total number of required ways. Therefore,
The total number of required ways of selecting the cards $= \dfrac{{{}^{52}{C_{17}} \times {}^{35}{C_{17}} \times {}^{18}{C_{17}} \times {}^1{C_1}}}{{3!}}$
Now, using the formula of combinations ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, we get
$\Rightarrow$ Total number of required ways of selecting the cards $= \dfrac{{\dfrac{{52!}}{{17!\left( {35} \right)!}} \times \dfrac{{35!}}{{17!\left( {18} \right)!}} \times \dfrac{{18!}}{{17!\left( 1 \right)}} \times 1}}{{3!}}$
Cancelling out the common terms, we get,
Total number of required ways of selecting the cards $= \dfrac{{52!}}{{{{\left( {17!} \right)}^3}3!}}$
Hence, a pack of 52 cards is divided into 4 sets, three of them having 17 cards each and the fourth just one card in $\dfrac{{52!}}{{{{\left( {17!} \right)}^3} \cdot 3!}}$ ways.
Therefore, option A is the correct answer.

Note: In this question, it is really important to keep in mind that we have to divide the total number of required ways by $3!$. This is because among the 4 sets, 3 sets are having 17 cards each, hence, these three sets will be considered as the groups having the same number of terms. Also, we should know that after the selection of 17 cards for the first set, we have to consider only the remaining cards for the next set, and so on. If we consider all 52 cards for the selection of the second and third set, then we will get the wrong answer.