x1 + x2 + x3 + x4 + x5 = 21, where xi , i = 1, 2, 3, 4, 5, is a nonnegative integer such that: 0≤ x1 ≤10 ,0≤ x2 ≤10 ,0≤ x3 ≤10 ,0≤ x4 ≤10 ,0≤ x5 ≤10 ,
Solved by Generating Function :
(1+ x + x2 +x3.....+ x10) (1+ x + x2 +x3.....+ x10)(1+ x + x2 +x3.....+ x10)(1+ x + x2 +x3.....+ x10)(1+ x + x2 +x3.....+ x10) = x21.
((1+ x + x2 +x3.....+ x10))5 = x21.
$\left ( \frac{1- x^{11}}{1-x} \right )^{5}$ = x21.
$\left ( 1- x^{11} \right )^{5}\left ( \frac{1}{1-x} \right )^{5}$ = x21.
$\left ( 1- x^{11} \right )^{5}\sum_{r=0}^{n}\binom{n+r-1}{r} x^{r} = x^{21}.$
$\left ( 1- x^{11} \right )^{5}\sum_{r=0}^{n}\binom{5+r-1}{r} x^{r} = x^{21}.$
$\left ( 1- x^{11} \right )^{5}\sum_{r=0}^{n}\binom{r+4}{r} x^{r} = x^{21}.$
Now we have to find term of x whose cofficient is 21.
$\left (\binom{5}{0} x^{0} - \binom{5}{1}x^{11} + \binom{5}{2}x^{22} - \binom{5}{3}x^{33}+ \binom{5}{4}x^{44} - \binom{5}{5}x^{55} \right )$ $\times \sum_{r=0}^{n}\binom{r+4}{r} x^{r} = x^{21}.$
$\left ( _{0}^{5}\textrm{C} \times _{21}^{25}\textrm{C}x^{21} \right ) - \left ( _{1}^{5}\textrm{C} \times _{10}^{14}\textrm{C}x^{21} \right ) = x^{21}$
$x^{21}\left ( 1 \times \left ( 25\times 23\times 22 \right ) \right ) - \left ( 5 \times\left ( 7\times 13\times 11 \right ) \right ) = x^{21}$
= 7645