Q.In how many ways can 5 letters be posted in 3 post boxes, if any number of letters can be posted in all of the three post boxes?
1. 5 C 3
2. 5 P 3
3. 5^3
4. 3^5
5. 2^5
Please help me with this sum
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- vinitrathi1
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- cans
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IMO D
first letter can be posted in any of the 3 boxes - 3
2nd----------------------------------------------- 3
3rd----------------------------------------------- 3
4th------------------------------------------------3
5th------------------------------------------------3
total = 3*3*3*3*3 = 3^5
first letter can be posted in any of the 3 boxes - 3
2nd----------------------------------------------- 3
3rd----------------------------------------------- 3
4th------------------------------------------------3
5th------------------------------------------------3
total = 3*3*3*3*3 = 3^5
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- Brent@GMATPrepNow
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The solution that cans gave is great.vinitrathi1 wrote: 1. 5 C 3
2. 5 P 3
I only want to add that, on the GMAT, students aren't required to know the combination (nCr) and permutation (nPr) notation.
The answer choices will given as either numbers or expressions.
Cheers,
Brent
- cans
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well write in proper form.I always get confused between the value of 'n' and 'r' in n^r. I went for 5^3.
Is there a easier way to avoid this confusion?
1st letter -3 ways
2nd - 3 ways
3rd - 3ways
4th-3 ways
5th-3 ways
thus 3*3*3*3*3 = 3^5 (this way you won't go for 5^3)
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- vinitrathi1
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I need to confirm 1 query -
Is it that in this question you have also given importance to the order in which the letter have been posted. As in, have you assumed that letters posted in this format (Letters 1 and 2 in 1st post box, letters 3 and 4 in 2nd post box and letter 5 in 3rd post box) is other than letter posted in this format(Letters 3 and 4 in 1st post box, letters 1 and 2 in 2nd post box and letter 5 in 3rd post box).
Please help!
Is it that in this question you have also given importance to the order in which the letter have been posted. As in, have you assumed that letters posted in this format (Letters 1 and 2 in 1st post box, letters 3 and 4 in 2nd post box and letter 5 in 3rd post box) is other than letter posted in this format(Letters 3 and 4 in 1st post box, letters 1 and 2 in 2nd post box and letter 5 in 3rd post box).
Please help!
- cans
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The letters can be in any order. (they are not numbered. I just said 1st,2nd letter to distinguish them.)vinitrathi1 wrote:I need to confirm 1 query -
Is it that in this question you have also given importance to the order in which the letter have been posted. As in, have you assumed that letters posted in this format (Letters 1 and 2 in 1st post box, letters 3 and 4 in 2nd post box and letter 5 in 3rd post box) is other than letter posted in this format(Letters 3 and 4 in 1st post box, letters 1 and 2 in 2nd post box and letter 5 in 3rd post box).
Please help!
and yes I assumed the above two cases you mentioned as different.
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- vinitrathi1
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Thanks for your help, I really appreciate!
Can I ask you another question? What really compels you to assume that they are different? (I assume that it is also not implied in the question)
Can I ask you another question? What really compels you to assume that they are different? (I assume that it is also not implied in the question)
- cans
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if all of them were identical, it would have been mentioned in the question
like: 5 identical letters were posted....
like: 5 identical letters were posted....
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- vinitrathi1
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- vinitrathi1
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Just curious, what would the answer be if the question was:
Q.In how many ways can 5 identical letters be posted in 3 post boxes, if any number of letters can be posted in all of the three post boxes?
Q.In how many ways can 5 identical letters be posted in 3 post boxes, if any number of letters can be posted in all of the three post boxes?
- cans
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let A,B,C order in which letters are on boxesvinitrathi1 wrote:Just curious, what would the answer be if the question was:
Q.In how many ways can 5 identical letters be posted in 3 post boxes, if any number of letters can be posted in all of the three post boxes?
5,0,0 = 3!/2! =3
4,0,1 = 3! =6
3,1,1 =3!/2! = 3
3,0,2 = 3! = 6
2,2,1= 3!/2! =3
total = 21
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