Question

How many solutions does the equation x_1 +x_2+x_3+x_4+x_5=21 have where x_1, x_2, x_3, x_4, and x_5 are nonnegative integers and x_1 >= 1?

Answer 1

Step-by-step explanation:

$x_{1} + x_{2} + x_{3} + x_{4} + x_{5} = 21$     (given)

Let us consider :

$x_{1}$ = $t_{1} + 1$

$x_{2}$ = $t_{2}$

$x_{3}$ = $t_{3}$

$x_{4}$  = $t_{4}$

$x_{5}$ = $t_{5}$

Now, by substituting the above considerations in the above equation, we get:

$t_{1} + 1 + t_{2} + t_{3} + t_{4} + t_{5} = 21$

$t_{1} + t_{2} + t_{3} + t_{4} + t_{5} = 20$

where,

$t_{i}$ $\geq$ 1

then it follows

n = 20

r = 4

then no. of solutions for the eqn = $_{r}^{n + r}\textrm{C}$

                                                      = $_{4}^{24}\textrm{C}$

                                                      = 10626

Answer :

no. of solutions for the eqn 10626