- #1

Ahmed Abdullah

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I have asked this question in yahoo answers. None of them have come out with the right answer. Amazing! It's such an easy problem.

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- Thread starter Ahmed Abdullah
- Start date

- #1

Ahmed Abdullah

- 203

- 3

I have asked this question in yahoo answers. None of them have come out with the right answer. Amazing! It's such an easy problem.

- #2

MathematicalPhysicist

Gold Member

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after that use the multiplication theorem.

wait, a second are you asking for help, or are you just giving us an example of yahoo stupidity?

- #3

Ahmed Abdullah

- 203

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my answer is 7!/2-6!/2=2160

I will be very glad if you carefully verify the answer.

Thnx

I will be very glad if you carefully verify the answer.

Thnx

- #4

MathematicalPhysicist

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combining them all by multiplication you get your answer, or so i think. (-:

- #5

HallsofIvy

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You don't give any reason why you subtract off 6!/2 so I can't comment on that. (My solution includes the possiblity that all 3 Ls will be together. I see nothing in the original question that excludes that.)

- #6

MathematicalPhysicist

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- #7

CRGreathouse

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You don't give any reason why you subtract off 6!/2 so I can't comment on that. (My solution includes the possiblity that all 3 Ls will be together. I see nothing in the original question that excludes that.)

Yoyu're double-counting when there are three Ls together, though, right? That is, you count (LL)L as different from L(LL), when they're actually the same.

- #8

Ahmed Abdullah

- 203

- 3

PARALLEL has 8 letters with one letter appearing 3 times and 1 letter appearing 2 times.

The letters can be arranged in 8!/(3!*2!) = 3360 ways.

If you want 2 Ls together, you can treat it is 1 single letter.

So it seems that the letters can be arranged in 7!/(2!) = 2520 ways.

But in this case you are treating the letter LL and L to be different. Yes they are different when they are in separate place (i.e when LL and L are not together). But they are not always different.

Consider the following cases

(1) P A R (LL) A E L and P A R L A E LL (two letter (LL) and L has just exchanged their position). These two permutations are not identical. so LL and L is different letters here.

But "PAR(LL)LAE" & "PARL(LL)AE" is identical though I have exchanged their position. So the letter (LL) and L are not same here. That infers the answer shouldn't be 7!/(2!*2!) = 1260 in which LL and L is considered identical in character.

**So I have to regard LL and L as identical letter when they are together and different when isolated. In how many permutations are they together? In 6!/2! permutation 3 Ls are together. **

If we could differ the third L from first two (LL) then there should be 7!/2! valid permutation (having 2 Ls together). But in these permutationS there are 2*6!/2! where 3 Ls are together and half of them is exactly same as the other half. So this half should be omitted to get the desired result.

So the number of valid permutation is 7!/2!-6!/2! = 2160

The letters can be arranged in 8!/(3!*2!) = 3360 ways.

If you want 2 Ls together, you can treat it is 1 single letter.

So it seems that the letters can be arranged in 7!/(2!) = 2520 ways.

But in this case you are treating the letter LL and L to be different. Yes they are different when they are in separate place (i.e when LL and L are not together). But they are not always different.

Consider the following cases

(1) P A R (LL) A E L and P A R L A E LL (two letter (LL) and L has just exchanged their position). These two permutations are not identical. so LL and L is different letters here.

But "PAR(LL)LAE" & "PARL(LL)AE" is identical though I have exchanged their position. So the letter (LL) and L are not same here. That infers the answer shouldn't be 7!/(2!*2!) = 1260 in which LL and L is considered identical in character.

If we could differ the third L from first two (LL) then there should be 7!/2! valid permutation (having 2 Ls together). But in these permutationS there are 2*6!/2! where 3 Ls are together and half of them is exactly same as the other half. So this half should be omitted to get the desired result.

So the number of valid permutation is 7!/2!-6!/2! = 2160

Last edited:

- #9

robert Ihnot

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I checked out some of these smaller cases, and the formula ever works in the event of ALLL, where there are no cases excluded. Here we have 4!/3! = 4, and actually 3!-2! = 4, the correct answer. I rewrite the problem as AuL, where u=LL, and further rewrite it as As, where s=LLL.

In AALLL, there are 5!/(2!3!) = 10 and the only excluded case is LALAL, so there are 9 cases involved. Using the formula: 4!/(2!)-3!/(2!) = 9, the correct answer. Now if we change that to ABLLL, lacking a double term in A, we have now 20 cases with 2 excluded, so correct anwer is 18. Changing this to ABuL, and ABs, and using formula, we get correct answer: 4!-3! = 24-6 = 18.

Take the case of BAALLL, equal 60 cases. Here we can put the two A's in the middle and have BLALAL, and move the B into anyone of the five other places giving 6 excluded cases. But in the form ALBLAL, we get only three new cases, ALBLAL, LBLAAL, LBLALA and similarly if we exchange the B for the other internal A, ALALBL, LAALBL, LALBLA. So 48 is the right answer. 5!/2-4!/2 = (120-24)/2 = 48.

In AALLL, there are 5!/(2!3!) = 10 and the only excluded case is LALAL, so there are 9 cases involved. Using the formula: 4!/(2!)-3!/(2!) = 9, the correct answer. Now if we change that to ABLLL, lacking a double term in A, we have now 20 cases with 2 excluded, so correct anwer is 18. Changing this to ABuL, and ABs, and using formula, we get correct answer: 4!-3! = 24-6 = 18.

Take the case of BAALLL, equal 60 cases. Here we can put the two A's in the middle and have BLALAL, and move the B into anyone of the five other places giving 6 excluded cases. But in the form ALBLAL, we get only three new cases, ALBLAL, LBLAAL, LBLALA and similarly if we exchange the B for the other internal A, ALALBL, LAALBL, LALBLA. So 48 is the right answer. 5!/2-4!/2 = (120-24)/2 = 48.

Last edited:

- #10

Ahmed Abdullah

- 203

- 3

I checked out some of these smaller cases, and the formula ever works in the event of ALLL, where there are no cases excluded. Here we have 4!/3! = 4, and actually 3!-2! = 4, the correct answer. I rewrite the problem as AuL, where u=LL, and further rewrite it as As, where s=LLL.

In AALLL, there are 5!/(2!3!) = 10 and the only excluded case is LALAL, so there are 9 cases involved. Using the formula: 4!/(2!)-3!/(2!) = 9, the correct answer. Now if we change that to ABLLL, lacking a double term in A, we have now 20 cases with 2 excluded, so correct anwer is 18. Changing this to ABuL, and ABs, and using formula, we get correct answer: 4!-3! = 24-6 = 18.

Take the case of BAALLL, equal 60 cases. Here we can put the two A's in the middle and have BLALAL, and move the B into anyone of the five other places giving 6 excluded cases. But in the form ALBLAL, we get only three new cases, ALBLAL, LBLAAL, LBLALA and similarly if we exchange the B for the other internal A, ALALBL, LAALBL, LALBLA. So 48 is the right answer. 5!/2-4!/2 = (120-24)/2 = 48.

Good examples!

- #11

robert Ihnot

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Thanks for the complement. I am glad you thought it was helpful.

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