Question

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

• A. $84$
• B. $56$
• C. $83$
• D. $112$

Answer 1

The correct option is B $56$

for a+b+c=n, number of triplets a,b,c is given by, (n+1)(n+2)/2

This can be derived as follows,

$consider(x+y+z{)}^n$

for 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 $(x+y+z{)}^n$

number of terms in $(x+y+z{)}^n$ is

$\frac{(n+1)(n+2)}{2}$

a+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+1

Then, a+b+c<=5

$a+b+c\le 5={\sum }_{r=0}^{5}n(a+b+c=r)$

Solving the above equation, we get number of triplets=56