A grandma has 30 presents to divide among her 10 grandchildren. If any division
ID: 2932873 • Letter: A
Question
A grandma has 30 presents to divide among her 10 grandchildren. If any division is equally likely, including that one lucky grandchild will get all the presents,
1) In how many way can she divide the presents?
2) If she decides to give 2 presents to each grandchild and then divide the remaining presents randomly, in how many ways can that be done?
3) Given that she has already given each grandchildren will remain with 2 gifts each, and all others will have at least 3 presents?
Could you please be kind enough and show which formula is applied, and how is calculated? Thank you!
Explanation / Answer
(Note: The presents here are considered identical. If this is not the case, find the answer in brackets.)
The number of ways to distribute r identical objects among n distinct containers is (n+r-1)C(r-1).
1. There are 30 presents that are to be distributed among 10 grandchildren.
This can be done in (30+10-1)C(10-1) ways
= 39C9 ways
= 211915132.
(If the presents are different, each present can have 10 choices. Thus 30 presents have 1030 different ways.)
2. After giving 2*10 = 20 presents, there are 10 presents remaining and they can be divided among 10 children in
(10 + 10 - 1) C (10 - 1)
= 19C9
= 92378 ways.
(If the presents are different, the 20 presents can be distributed in 20C2 * 18C2 * ...... 4C2 * 2C2 ways and the remaining 10 in 1010 different ways.)
3. I suppose two should be distributed among 5 children and we need to find the probability that the remaining will get atleast three presents.
By Baye's theorem, the number is P(two should be distributed among 5 children AND the remaining will get atleast three presents) / P(two should be distributed among 5 children)
Since each child should get 2, 2*5 = 10 presents should be distributed. This can be done in (10+10-1)C(10-1) = 19C9 ways.
If 5 have 2 each and 5 others have 3 each, we will have 5 presents left, which can be distributed among the 5 children in (5+5-1)C(5-1) ways.
The answer should be (5+5-1)C(5-1) / 19C9.
(Unless there is an additional condition, the 5C3 should not be there.)
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