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Folks i think if someone can highlight the difference

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SVP
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Folks i think if someone can highlight the difference [#permalink] New post 06 Aug 2006, 02:47
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Folks i think if someone can highlight the difference between those two problems , will make concepts more clear .. at least for me.

I have a major problem understanding the counting principles .

1st one:

In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys?

20
30
50
600
Correct Answer is 50

2nd:

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) 5C3
(2) 5P3
(3) 53
(4) 35
Correct Answer - (4)
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 [#permalink] New post 26 Dec 2006, 21:03
The difference is simple; in the first question there can't be any empty boxes, whilst in the second there can be.

BTW, the answer for the second question is 3^5 and not 35
Senior Manager
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Re: counting and permutations [#permalink] New post 27 Dec 2006, 02:52
yezz wrote:
1st one:

In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys?

20
30
50
600
Correct Answer is 50




5 different toys and 3 identical boxes

none of the boxes is to be empty, so there are two possibilities
(1) one box contains 3 toys and two boxes contain 1 toy each
Distribution - 3 :1:1
No. of ways = 5c3 * 2c1 * 1c1 = 20

(2) One box contains 1 toy and other two boxes contain 2 toys each
Distribution- 1:2:2
No. of ways= 5c1 * 4c2 * 2c2 = 30

Total no. of ways = 20+30 = 50
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Re: counting and permutations [#permalink] New post 27 Dec 2006, 02:57
yezz wrote:
2nd:

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) 5C3
(2) 5P3
(3) 53
(4) 35
Correct Answer - (4)


Each letter can be posted in any 3 post boxes
so no of ways = 3*3*3*3*3 = 3 ^ 5
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 [#permalink] New post 31 Dec 2006, 10:10
AK,
For the 1st case shouldn't the answer be 3*(20) + 3*(30) = 150?
You considered cases 3:1:1 and 1:1:2. However how about cases 1:3:1 and 1:1:3 and 1:2:1 and 2:1:1

thanks.
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 [#permalink] New post 01 Jan 2007, 06:25
Since the boxes are identical, you need not multiply by 3
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 [#permalink] New post 01 Jan 2007, 17:14
compuser1978 wrote:
AK,
For the 1st case shouldn't the answer be 3*(20) + 3*(30) = 150?
You considered cases 3:1:1 and 1:1:2. However how about cases 1:3:1 and 1:1:3 and 1:2:1 and 2:1:1

thanks.


since boxes are similar : 3 :1:1 / 1:3:1 / 1:1:3 ..all are same
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 [#permalink] New post 05 Jan 2007, 16:24
1) Items A, B, C, D, E. Ways to distribute them in the 3 identical boxes: 1 2 2, 1 3 1.

1 2 2:
#ways to pick item for 1st box = 5C1
#ways to pick 2 items for 2st box = 4C2
#ways to pick 2 items for 3st box = 2C2
Total #ways = 5C1 * 4C2 * 2C2 = 5 * 6 * 1 = 30

In a same fashion for 1 3 1:
Total #ways = 5C1 * 4C3 * 1C1 = 5 * 4 * 1 = 20

Total = 30 + 20 = 50

2) as mentioned above: 3 * 3 * 3 * 3 * 3 = 3^5.
  [#permalink] 05 Jan 2007, 16:24
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Folks i think if someone can highlight the difference

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