(a) The number of solutions of ${x_1} + {x_2} + … + {x_k} = n$

= Coefficient of tn in $(t + {t^2} + {t^3} + …)({t^2} + {t^3} + …)$

                                                                      …..$({t^k} + {t^{k + 1}} + ….)$

= Coefficient of ${t^n}$in ${t^{1 + 2 + … + k}}{(1 + t + {t^2} + …)^k}$

But $1 + 2 + … + k = \frac{1}{2}k(k + 1) = r$(say)

and $1 + t + {t^2} + …. = \frac{1}{{(1 – t)}}$

Thus number of required solutions

= Coefficient of ${t^{n – r}}$ in ${(1 – t)^{ – k}}$

=Coefficient of ${t^{n – r}}$ in

$[1 + {\,^k}{C_1}t + {\,^{k + 1}}{C_2}{t^2} + {\,^{k + 2}}{C_3}{t^3} + …]$

$ = {\,^{k + n – r – 1}}{C_{k – 1}}$ ${x^2} – (p + \alpha )x + p\alpha + q\beta = 0$, (where

$m = k + n – r – 1$)

$ = k + n – 1 – \frac{1}{2}k(k + 1) = \frac{1}{2}[2k + 2n – 2 – {k^2} – k]$

$ = \frac{1}{2}(2n – {k^2} + k – 2)$