Thread: Counting principles, three letter sequence w/o repeated letters, containing e or f

1. Counting principles, three letter sequence w/o repeated letters, containing e or f

How many three-letter sequences without repeated letters can be made using a,b,c,d,e,f in which either e or f (or both) is used?

I looked at the total sequences posible with all letters:
6*5*4=120 this is w/o replacement

The I looked at the amount of sequences without e or f:
4*3*2=24

Then I substract the amount of sequences without e or f from the total amount of possible sequences:
(6*5*4)-(4*3*2)=96 amount of sequences with e or f

But now I want to find the amount of sequences with e or f directly.
Here I have a problem because it is e or f.

If I would look for "without repetition and containing the letter g," I would do
g _ _ 5 * 4 = 20
NOg g _ 5 * 4 = 20
NOg NOg g 5 * 4 = 20 and then add it up 20 + 20 + 20 = 60

How can I do this with g or f?

2. I think you have to use an inclusion-exclusion argument. The number of triples containing an e is 60. So is the number containing an f. But if you just add those two, you are double-counting the triples that contain both e and f. So you need to count how many there are of those (in fact, 24), and subtract that number. Then that gives you the answer 60 + 60 – 24 = 96 for the number containing either e or f (or both), in agreement with the previous method.

3. Thank you for your answer. But I was looking for a solution withour a substruction. The book gives the answer (2*3*4*3) + (3*2*4) = 96 and I do not understand how to get this answer.

4. Originally Posted by DBA
Thank you for your answer. But I was looking for a solution withour a substruction. The book gives the answer (2*3*4*3) + (3*2*4) = 96 and I do not understand how to get this answer.
If exactly one of e, f occurs then there are 2 choices for which of them it is, 3 choices for the position in which it occurs, 4 choices for the letter in the first vacant position, and 3 choices for the letter in the remaining vacant position: 2*3*4*3 = 72 possibilities.

If both of e, f occur then there are 3 choices for the position in which e occurs, 2 choices for the position in which f occurs, and 4 choices for the letter to go in the remaining vacant position: 3*2*4 = 24 possibilities.