Question

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

• A. 84
• B. 56
• C. 83
• D. none of these

Answer 1

Answer: (B)

56

Now since a,b,c are positive numbers, and $a+b+c\le 8$
Hence
$a+b+c=8$
$a+b+c=7$
$a+b+c=6$
:
:
:
$a+b+c=3$
Hence
If we fix a=1 and b=1 then number of ways in which we get $a+b+c=8$
Is $\frac{6(6+1)}{2}=21$
Similarly if $a=2$ $b=1$ number of ways of filling c is 5.
Hence required permutation
$\frac{5(5+1)}{2}=15$
And so on.
Hence
${}^7{C}_2+{}^6{C}_2+{}^5{C}_2+{}^4{C}_2+{}^3{C}_2+1$
$=21+15+10+6+3$
$=55+1$
$=56$
Hence answer is 56.
In the above ordered pairs, all the triplets for $a+b+c\le 8$ are covered.