    Question

# If a, b, c are positive integers such that a+b+c≤8 then the number of possible 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
none of these
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Solution

## The correct option is C 56Now since a,b,c are positive numbers, and a+b+c≤8Hencea+b+c=8a+b+c=7a+b+c=6:::a+b+c=3HenceIf we fix a=1 and b=1 then number of ways in which we get a+b+c=8Is 6(6+1)2=21Similarly if a=2 b=1 number of ways of filling c is 5.Hence required permutation5(5+1)2=15And so on.Hence7C2+6C2+5C2+4C2+3C2+1=21+15+10+6+3=55+1=56Hence answer is 56.In the above ordered pairs, all the triplets for a+b+c≤8 are covered.  Suggest Corrections  0      Similar questions  Related Videos   Let's Solve a Problem
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