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yezz
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yezz

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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compuser1978
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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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Since the boxes are identical, you need not multiply by 3
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AK
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compuser1978
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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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.



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