Question

# If a,b,c are positive integers such that a+b+c≤8, then the number of positive values of the ordered triplet (a,b,c) is:

A
84
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B
56
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C
83
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D
112
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Solution

## The correct option is B 56for a+b+c=n, number of triplets a,b,c is given by, (n+1)(n+2)/2This can be derived as follows,consider (x+y+z)nfor all the terms in the equation, sum of powers=n therefore the number triplets such that a+b+c=n is equal to the number of terms in (x+y+z)nnumber of terms in (x+y+z)n is (n+1)(n+2)2a+b+c<=8, a,b,c >0 minimum value of a,b,c=1. so replace a by a+1,b by b+!, c by c+1Then, a+b+c<=5a+b+c≤5=∑5r=0n(a+b+c=r)Solving the above equation, we get number of triplets=56

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