# Permutation combination

1. Nov 1, 2014

### Robert Houdart

1. The problem statement, all variables and given/known data
In how many ways can one choose 6 cards from a normal deck of cards so as to have all suits present?

2. Relevant equations

3. The attempt at a solution
4 different cards can be chosen in 13*13*13*13 ways. Now we have to choose 2 remaining cards from 48 cards. This can be done in 48C2 ways.. Therefore, total number of ways will be 13*13*13*13*47*24

However, the book says 13*13*13*13*48*47 (I tend to disagree since we have to choose the remaining 2 cards rather than arrange them) am i right??

2. Nov 1, 2014

### PeroK

Both answers look wrong to me! I'm assuming that the order the cards are drawn does not matter. To see this just consider the Hearts.

You have 13 choices of Heart - let's say we have the 2 of Hearts. But, if the first extra card is a Heart (5, say), then this will represent the same situation with these the other way round.

To do it correctly, you have to consider the cases where you have: 3 of 1 suit and 1 each of the others; and, 2 each of 2 suits and 1 each of the others.

3. Nov 1, 2014

### Orodruin

Staff Emeritus
I agree with PeroK, both of your mentioned possibilities are definitely wrong. There is no requirement that the first four cards you draw need to be of different suits and several of the options you are counting are going to be equivalent.

4. Nov 1, 2014

### Ray Vickson

Your answer is 32,216,808, the book's answer is 64,433,616, but the correct answer is between 6,000,000 and 12,000,000; I won't say now exactly what it is. Reasons for the descrepancy has already been explained informally by others.

5. Nov 1, 2014

### PeroK

The OP seems to have gone. In any case, I wondered if it could be done using the original method and, if one is careful, it can be done:

$(13)^4(48x)(36y + 11z)$

Where x, y and z need to be chosen to take care of any duplicates.

6. Nov 1, 2014

### Orodruin

Staff Emeritus
Just to prevent any future needs for warnings: The OP has only been away for a few hours. It is far too early to start discussing the solution.

7. Nov 1, 2014

### haruspex

Not sure how you are getting that, but it doesn't fit with the answer I get. If you want to discuss further, better use a private conversation.

8. Nov 2, 2014

### Robert Houdart

Let us take suits as A, B, C, D
Now the different case which arises are
3 A 1 B 1 C 1D
1 A 3 B 1 C 1D
1 A 1 B 3C 1D
1 A 1 B 1C, 3 D
2 A 2 B 1C, 1D
2 A 1 B 2C 1D
2 A 1 B 1C 2 D
1 A 2 B 2C 1D
1 A 2 B 1C 2 D
1 A 1 B 2C 2 D

Therefore required number of ways will be 13*13*13 (4* 13c3) + 13*13*6 (13c2 * 13c2) = 6217510
So am I right this time??

9. Nov 2, 2014

### Ray Vickson

10. Nov 2, 2014

### Robert Houdart

I guess 8682544 is the correct answer.

11. Nov 2, 2014

### Orodruin

Staff Emeritus
Correct. There is also a more straight forward way of counting the possible different cases than listing them and risking to miss one. Can you figure it out?

12. Nov 2, 2014

Yes, it is.