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Question

The number of ordered triplets of positive integers which are solutions of the equation x+y+z=100 is?


A

6005

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B

4851

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C

5081

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D

None of these

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Solution

The correct option is B

4851


Explanation for the correct option:

Find the required number of ordered triplets.

An equation x+y+z=100 is given.

Since the sum n of all variables is 100 and the number r of variables is 3.

So, the total number of triplets is given by Cr-1n-1.

Cr-1n-1=C3-1100-1Cr-1n-1=C299Cr-1n-1=99·982Cr-1n-1=4851

Therefore, The number of ordered triplets of positive integers which are solutions of the equation x+y+z=100 is 4851.

Hence, option B is the correct answer.


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