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29 Apr 2021, 06:15
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
52% (02:02) correct
48%
(02:06)
wrong
based on 107
sessions
History
Date
Time
Result
Not Attempted Yet
The number of ordered triplets of positive integers which are the solutions of the equation x + y + z = 100 is
(A) 4851
(B) 5081
(C) 6005
(D) 6151
(E) 6273
(A) 4851
(B) 5081
(C) 6005
(D) 6151
(E) 6273
29 Apr 2021, 06:57
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Bunuel
This is the same as distributing 100 identical toffees among 3 children. However, each child must get at least 1 toffee.
So, let us give 1 toffee to each child. That leaves us with 97 toffees to be distributed among 3 children.
Let the toffees be denoted as T.
There are 97 Ts in a line.
To distribute these among 3 children, we need to make 2 partitions (shown as p in the diagram below):
T T T T ... p T T p T T T T ...
Depending on where we place the 'p', we will get different distributions for the 3 children.
However, as we keep doing this, we keep making different words.
Thus, the number of distributions is the same as the number of words formed.
Total words formed by 97 T and 2 p
= 99!/(97!*2!)
= 99*98/2 = 99*49 = 4851
Answer A
Posted from my mobile device
General Discussion
08 May 2021, 01:56
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sujoykrdatta
sujoykrdatta,
Nice soln.
However can you explain how to do the below question with your method
How many ordered triplets (a, b, c), where a, b, and c are non-negative integers, are there such a + b + c = 10 ?
Here each student need NOT have at least one .
https://gmatclub.com/forum/how-many-ord ... 23382.html
Thank you.
