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Folks i think if someone can highlight the difference

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SVP
Joined: 05 Jul 2006
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Folks i think if someone can highlight the difference [#permalink]

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06 Aug 2006, 03:47
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

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
Manager
Joined: 03 Dec 2006
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26 Dec 2006, 22:03
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
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Joined: 19 Jul 2006
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27 Dec 2006, 03:52
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

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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27 Dec 2006, 03:57
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

Each letter can be posted in any 3 post boxes
so no of ways = 3*3*3*3*3 = 3 ^ 5
Intern
Joined: 22 Nov 2006
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31 Dec 2006, 11:10
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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01 Jan 2007, 07:25
Since the boxes are identical, you need not multiply by 3
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01 Jan 2007, 18:14
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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05 Jan 2007, 17:24
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.
05 Jan 2007, 17:24
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