Perms and Combs Help (1 Viewer)

[bujolover]

Hi guys,
I have some (more!) questions; this time they're related to permutations and combinations. I have a very strong suspicion (read: almost certainty) that the given answers for these questions are incorrect, but I need confirmation from you guys...

How many different arrangements can be made from the letters of the word EXERCISE if:
i. there are no restrictions?

The given answer for this question is 8! = 40320. But don't you need to divide that by 3! to get 6720, because there are 3 identical E's?

ii. the letters C and R are at the ends?

The given answer for this question is 6!.6 = 4320, but I divided 6! by 3! (6), rather than multiply the two. It doesn't make sense to multiply the two to me anyway, but correct me if I'm wrong.

I'm sorry if these are obviously wrong, but it's sort of an OCD thing- I won't be satisfied until I get confirmation.

Thanks in advance!

P.S. It doesn't matter if you post your answer to only one of these questions. Any help at all is help.

 

[pikachu975]

I think you're right for part i since they are identical letters.

For part ii you do 1x1 for the C and R and 6!/3! due to identical E's which is what I think you did so the answers provided are wrong I think.

Edit: 2*6!/3!

 

[sida1049]

I'm not particularly good at permutations and combinations, but...

For part (ii),

I got 2.6!/(3!).

6! because of the letters in the middle. Divide by 3! because of the 3 identical E's. Multiply by 2 because we can have the word beginning with C and ending with R, or beginning with R and ending with C.

 

[tsoliman1]

Hi guys,
I have some (more!) questions; this time they're related to permutations and combinations. I have a very strong suspicion (read: almost certainty) that the given answers for these questions are incorrect, but I need confirmation from you guys...

How many different arrangements can be made from the letters of the word EXERCISE if:
i. there are no restrictions?

The given answer for this question is 8! = 40320. But don't you need to divide that by 3! to get 6720, because there are 3 identical E's?

ii. the letters C and R are at the ends?

The given answer for this question is 6!.6 = 4320, but I divided 6! by 3! (6), rather than multiply the two. It doesn't make sense to multiply the two to me anyway, but correct me if I'm wrong.

I'm sorry if these are obviously wrong, but it's sort of an OCD thing- I won't be satisfied until I get confirmation.

Thanks in advance!

P.S. It doesn't matter if you post your answer to only one of these questions. Any help at all is help.

For ii)
If that's the exact wording of the question, then i think it would be 6!/3! as mentioned but then multiplied by 2 to account for the C and R switching places

 

[bujolover]

I think you're right for part i since they are identical letters.

For part ii you do 1x1 for the C and R and 6!/3! due to identical E's which is what I think you did so the answers provided are wrong I think.

Thanks for part i. But for part ii, as the other 2 guys realised, and I just realised, you have to multiply that answer by 2...so I guess the answer is 240. Either way, both given answers are wrong, so thanks for confirming that.

 

[bujolover]

For ii)
If that's the exact wording of the question, then i think it would be 6!/3! as mentioned but then multiplied by 2 to account for the C and R switching places

I'm not particularly good at permutations and combinations, but...

For part (ii),

I got 2.6!/(3!).

6! because of the letters in the middle. Divide by 3! because of the 3 identical E's. Multiply by 2 because we can have the word beginning with C and ending with R, or beginning with R and ending with C.

Thanks, guys. I forgot that there are 2! ways of arranging the C and the R, so thanks for reminding me about that (see, this is why I ask even the most seemingly trivial questions ), and also for confirming that both given answers are wrong.