Question
If $$x, y, z$$ are integers and $$x \geqslant 0,y \geqslant 1,z \geqslant 2,x + y + z = 15$$ then the number of values of the ordered triplet $$(x, y, z)$$ is
A.
$$91$$
B.
$$455$$
C.
$$^{17}{C_{15}}$$
D.
None of these
Answer :
$$91$$
Solution :
Let $$y = p + 1, z = q + 2.$$ Then $$x \geqslant 0,p \geqslant 0,q \geqslant 0$$ and $$x + y + z = 15$$ implies $$x + p + q = 12.$$
∴ the required number of values of $$(x, y, z)$$ and hence of $$(x, p, q)$$
= the number of non-negative integral solutions of $$(x + p + q = 12)$$
= co-efficient of $${x^{12}}\,{\text{in }}{\left( {{x^0} + {x^1} + {x^2} + .....} \right)^3}$$
= co-efficient of $${x^{12}}\,{\text{in }}{\left( {1 - x} \right)^{ - 3}}$$
= co-efficient of $${x^{12}}\,{\text{in }}\left\{ {^2{C_0} + {\,^3}{C_1}x + {\,^4}{C_2}x + .....} \right\}$$
$$ = {\,^{14}}{C_{12}} = \frac{{14!}}{{12!\,\,2!}} = \frac{{14 \times 13}}{2} = 91.$$