Question

The number of ways can a pack of 52 cards be divided into 4 sets, three of them having 17 cards each and fourth just one card is
(a) \[\dfrac{52!}{{{\left( 17! \right)}^{3}}}\]
(b) \[\dfrac{52!}{3\cdot {{\left( 17! \right)}^{3}}}\]
(c) \[\dfrac{52!}{3!{{\left( 17! \right)}^{3}}}\]
(d) \[\dfrac{52!}{{{\left( 3! \right)}^{3}}17!}\]

Answer 1

Hint: In this question, we first need to select 17 cards out of 52 cards which can be done by using the combinations formula given by \[{}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}\]. Then we need to again select 17 cards from the remaining 35 cards using the combination formula. Now, again select another 17 cards from 18 cards which can be done using the \[{}^{n}{{C}_{r}}\] formula. Then we will be left with only 1 card which can be done in only 1 way. Now, multiplying all these values gives the result.

Complete step-by-step answer:
Now, from the given question we have a pack of 52 cards
As we already know that selecting a given number of things without reference to the order of the things is called a combination.
The number of combinations of n different things taken r at a time is
\[{}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}\]
Now, we need to divide 52 cards into 3 sets of 17 cards and 1 set of 1 card.
Let us first select 17 cards out of 52 cards using the combination formula
\[\Rightarrow {}^{52}{{C}_{17}}\]
Now, this can be further written as
\[\Rightarrow \dfrac{52!}{35!17!}\]
Here, as we selected 17 cards out of 52 now we are left with 35 cards which can be further divided into the respective sets.
Now, we have 35 cards out of which we need to select another set of 17 cards
\[\Rightarrow {}^{35}{{C}_{17}}\]
Now, using the formula we can also write it as
\[\Rightarrow \dfrac{35!}{18!17!}\]
Now, we are left with 18 cards in which we need to select 17 cards out of it
\[\Rightarrow {}^{18}{{C}_{17}}\]
Now, this can be further written as
\[\Rightarrow \dfrac{18!}{1!17!}\]
Now, we have divided 3 sets with 17 cards and left with only one card which can be get into the set in only one way.
Now. the total number of ways in which 52 cards can be divided into 4 sets in which 3 sets having 17 cards and 1 card in the other set is given by
\[\Rightarrow \dfrac{52!}{35!17!}\times \dfrac{35!}{18!17!}\times \dfrac{18!}{1!17!}\times 1\]
Now, on cancelling out the common terms we get,
\[\Rightarrow \dfrac{52!}{17!}\times \dfrac{1}{17!}\times \dfrac{1}{1!17!}\times 1\]
Now, this can be further written in the simplified form as
\[\Rightarrow \dfrac{52!}{{{\left( 17! \right)}^{3}}}\]
Hence, the correct option is (a).

Note:Instead of selecting 17 cards first three times and dividing them into 3 sets we can also solve by first selecting the one card as one set and then select 17 cards from the remaining cards and follow similarly. Both the methods give the same result.It is important to note that we just need to select the cards to divide them into sets but not arrange. So we use the combination formula to solve it, not the permutation formula. Because doing the other way changes the result which are also mentioned in the options.