# Homework Help: Permutation Question

1. Apr 2, 2012

### Michael_Light

1. The problem statement, all variables and given/known data

I have 6 cards with the digits 1,2,2,3,4 and 5 written on each of them. In how many ways can I arrange 3 out of the 6 cards in a row?

2. Relevant equations

3. The attempt at a solution

Case 1: Only one card with digit '2' two is taken, so 5P3 =60

Case 2: Both cards with digit '2' are used. So 4 x (3!/2) =12

Total: 60+12 =72

My question is, do we need to consider the third case where no cards with digit '2' is taken, which gives additional arrangement of 4! =24?

I really confused.. Please guide me.. Or do we have other method to solve this..? Thanks..

2. Apr 2, 2012

### Ray Vickson

You say the numbers are written on "each of them"; in standard English, that means that each of the six cards has those six numbers on them. However, some of your later statements indicate this is not the case.

Assuming that each card has only ONE number on it, it is easier to consider the two 2's as being temporarily distinguished, by calling them 2a and 2b. After counting, you can erase the 'a' and 'b' and so divide by two all counts having one or both 2's in the arrangement.

RGV

Last edited: Apr 2, 2012
3. Apr 2, 2012

### HallsofIvy

First imagine that the two cards with "2" on them have "2" and "2+" so that they are "distinguishable". There are 6 distinguishable cards. How many linear arrangements of 3 out 6 things are there?

Now, there three possibilities to take into account. It might be that neither of the "2"s is in those three cards. In how many ways can that take place?

It might be that exactly one "2" is in the three cards. In how many ways can that be done?

It might be that both "2"s are in the three cards. In how many ways can that be done?

You will want to divide the first number by the other three.

Last edited by a moderator: Apr 2, 2012
4. Apr 2, 2012

### Michael_Light

I still have doubts, but this is how i interpret what you guys wrote...

Assuming the '2' is distinguishable, then it will have 63 =120 arrangement.

If only one card with digit '2' two is taken, it will be 5P3 =60 arrangement.

If both cards with digit '2' are taken, it will be 4 x (3!/2) =12 arrangement.

If no cards with digit '2' is taken, then 4! =24 arrangement.

Am i in the right track and how should i proceed? Thank you..

Last edited: Apr 2, 2012
5. Apr 2, 2012

### RoshanBBQ

We can use a method similar to yours, I believe, if we modify it slightly. And yes, you need to consider when no cards with a 2 is chosen. Let me go over the procedure for one case.
Case 1: one 2
If one 2 is chosen, you can think of a single slot already taken. Hence, you can first find the number of ways to order 2 from a set of 4 (since 4 numbers remain after removing the two 2s and you must fill 2 slots since 1 is taken by a 2). But there is an additional step. You must multiply by 3, because your 2 could be in slot 1, slot2, or slot3. Think about it this way, nPr(4,2) = 12 and they are
13 31 14 41 15 51 34 43 35 53 45 54

You then can put the 2 at the end
213 231 214 241 215 251 234 243 235 253 245 254
Middle
...
Or end
...
So it is 3 times the result.

We can do a smaller example to explore this logic:
Order 2 of 4 where you can pick from {1,2,2,4}.
If you want to know how many ways you can do it, you can add the number of ways for having a single 2, no 2, or both 2s.
Case 1: one 2
12
21
32
23
So 4. Let us apply the logic above to compute this now: nPr(2,1)*2=4. So we have to choose 1 (since 1 slot remains unfilled) from a selection of 2 (since 1 and 4 are all that remain after removing 2) where order matters and then multiply it by two, because the 2 can be in slot 1 or slot2.

Just a side note, in general, the amount you multiply by is nCr(slots,indistinguishable objects taken to be already chosen). To bring concreteness, let us try to compute the 3 we already thought out in the first example. We can think of choosing from 3 positions 1 slot. Order doesn't matter, because the 2s being placed are indistinguishable. So nCr(3,1) = 3. If we had, instead, two 2s known to be placed in a selection of 4 cards, we'd have to multiply by
22xx
2x2x
2xx2
x22x
x2x2
xx22
6. So nCr(4,2) = 6

Case 2: no 2s
Well, you then choose 2 out of the remaining set: nPr(2,2)=2. Verification:
14
41

Case 3: both 2s
Well, in this simple example, there is only 1 way to have 22.
1

Total: 1 + 4 + 2 = 7.

Last edited: Apr 2, 2012
6. Apr 2, 2012

### Ray Vickson

Your case 1 is incorrect: the P(5,3) = 60 arrangements include some in which the '2' is not chosen at all. So, you must instead do what others have suggested: N(one 2) = 36, obtained as follows: first, choose the two non-2 cards; there are P(4,2) = 4*3 = 12 such permutations. Now insert a '2' into one of the three available positions. That gives a total of 3*12 = 36 arrangements.

For the number having no '2', just use P(4,3) = 4*3*2 = 24.

Finally, for the number having two 2s: there are 4 choices for the non-2, then there are 3 places to put that non-2, for a total of 4*3 = 12 arrangements.

RGV