in how many ways can a pack of 52 cards be divided into 4 sets three of them having 16 cards each and the fourth just 4 cards ?

Options

A. `16 ! Xx 52 !`
B. `((52!))/((16!)^3)`
C. `(52 !)/((3!)^(16))`
D. `(52!)/((16)^3 xx(3!))`

Answer

To solve the problem of dividing a pack of 52 cards into 4 sets, where 3 sets contain 16 cards each and the 4th set contains 4 cards, we can follow these steps:

### Step-by-Step Solution:

1. **Choose Cards for the First Set**:
We start by selecting 16 cards from the total of 52 cards. The number of ways to choose 16 cards from 52 is given by the combination formula:
\[ \binom{52}{16} \]

2. **Choose Cards for the Second Set**:
After selecting the first set of 16 cards, we have 36 cards remaining. Now, we need to choose another 16 cards for the second set. The number of ways to choose 16 cards from the remaining 36 is:
\[ \binom{36}{16} \]

3. **Choose Cards for the Third Set**:
After selecting the first two sets, we have 20 cards left. We now choose 16 cards for the third set from these 20 cards. The number of ways to choose 16 cards from 20 is:
\[ \binom{20}{16} \]

4. **Choose Cards for the Fourth Set**:
Finally, we have 4 cards left, which will automatically form the fourth set. The number of ways to choose all 4 cards from the remaining 4 is:
\[ \binom{4}{4} = 1 \]

5. **Combine the Choices**:
The total number of ways to choose the cards for all four sets is the product of the combinations calculated above:
\[ \text{Total Ways} = \binom{52}{16} \times \binom{36}{16} \times \binom{20}{16} \times \binom{4}{4} \]

6. **Account for Identical Sets**:
Since the first three sets are identical in size (16 cards each), we must divide by the number of ways to arrange these 3 identical sets, which is \(3!\):
\[ \text{Final Answer} = \frac{\binom{52}{16} \times \binom{36}{16} \times \binom{20}{16}}{3!} \]

### Final Formula:
Putting it all together, the final formula for the number of ways to divide the pack of 52 cards into the specified sets is:
\[ \text{Final Answer} = \frac{\binom{52}{16} \times \binom{36}{16} \times \binom{20}{16}}{6} \]