Question

The number of ways in which 52 playing cards can be divided into four sets, three of them having 17 cards each and fourth are having just one card is:
A. $\dfrac{{52!}}{{{{\left( {17!} \right)}^3}}}$
B. $\dfrac{{52!}}{{{{\left( {17!} \right)}^3}3!}}$
C. $\dfrac{{51!}}{{{{\left( {17!} \right)}^3}}}$
D. $\dfrac{{51!}}{{{{\left( {17!} \right)}^3}3!}}$

Answer 1

Here, we will use combination whose formula is given as $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ to determine the number of ways. In order to find the required number of ways first, distribute 17 cards from 52 cards, then 17 cards from remaining cards, that is 35 cards, and again 17 cards to third set from 18 left cards and so on. We will also divide by 3! As the order of the sets does not matter.

Complete step by step solution:
We are given that a pack has 52 cards.
We have to divide 52 cards into 4 sets, such that there are 17 cards in 3 sets and 1 card in one set.
Let the number of ways of doing so be ${N_1}$.
First, we will give 17 cards from 52 cards to one person.
Then the number of ways of doing it, say ${N_2}$ be $^{52}{C_{17}} = \dfrac{{52!}}{{17!\left( {52 - 17} \right)!}} = \dfrac{{52!}}{{17!35!}}$
Next, we are left with 35 cards and we will be distributing 17 cards from it.
The number of ways of doing it, ${N_3}$ be $^{35}{C_{17}} = \dfrac{{35!}}{{17!\left( {35 - 17} \right)!}} = \dfrac{{35!}}{{17!18!}}$
Similarly, we will select 17 cards from the left cards.
Hence, the number of ways of doing above is ${N_4}$ which is $^{18}{C_{17}} = \dfrac{{18!}}{{17!\left( {18 - 17} \right)!}} = \dfrac{{18!}}{{17!1!}}$
And now we have to select one card from 1 card for fourth set.
Therefore, the number of ways of doing it is ${N_5}{ = ^1}{C_1} = \dfrac{{1!}}{{1!1!}}$
The order will not matter =, therefore, we will divide by 3! for three same groups.
Hence, the value of ${N_1}$ is $\dfrac{{{N_2} \times {N_3} \times {N_4} \times {N_5}}}{{3!}}$
Then, ${N_1} = \dfrac{{52!}}{{17!35!}} \times \dfrac{{35!}}{{17!18!}} \times \dfrac{{18!}}{{17!1!}} \times \dfrac{{1!}}{{1!0!}} \times \dfrac{1}{{3!}}$
On solving the above expression, we will get,
${N_1} = \dfrac{{52!}}{{17!35!}} \times \dfrac{{35!}}{{17!18!}} \times \dfrac{{18!}}{{17!1!}} \times \dfrac{{1!}}{{1!0!}} \times \dfrac{1}{{3!}} \\\ {N_1} = \dfrac{{52!}}{{{{\left( {17!} \right)}^3}\left( {3!} \right)}} \\\$

Hence, option B is correct.

Note:
Students generally forget dividing by 3! after multiplying all the number of ways. Here, we have to take care that the order of the set does not matter. And there are 3 sets of 17 cards. We have to distribute cards irrespective of the order. Hence, we will divide by 3! to get the final answer.