How-many-3-digits-number-such-that-sum-of-its-digits-is-11-

Question Number 160909 by naka3546 last updated on 09/Dec/21

H o w m a n y 3 - d i g i t s n u m b e r s u c h t h a t s u m o f i t s d i g i t s i s 11 ?

Answered by mr W last updated on 09/Dec/21

M E T H O D 1

(x + x^{2} + x^{3} + \dots x^{9}) {(1 + x + x^{2} + \dots + x^{9})}^{2}

= \frac{x (1 - x^{9}) {(1 - x^{10})}^{2}}{{(1 - x)}^{3}}

= x (1 - x^{9} - 2 x^{10} + 2 x^{19} + x^{20} - x^{29}) \underset{k = 0}{\sum^{\infty}} C_{2}^{k + 2} x^{k}

c o e f . o f x^{11} i s

C_{2}^{12} - C_{2}^{3} - 2 \times C_{2}^{2} = 66 - 3 - 2 = 61

t h a t m e a n s t h e r e a r e 61 s u c h 3 - d i g i t

n u m b e r s .

Commented by Tawa11 last updated on 09/Dec/21

Great sir .

Answered by mr W last updated on 09/Dec/21

M E T H O D 2

a b c w i t h a + b + c = 11

a = 1 : b + c = 10 \Rightarrow 9 n u m b e r s

a = 2 : b + c = 9 \Rightarrow 10 n u m b e r s

a = 3 : b + c = 8 \Rightarrow 9 n u m b e r s

a = 4 : b + c = 7 \Rightarrow 8 n u m b e r s

\dots

a = 9 : b + c = 2 \Rightarrow 3 n u m b e r s

t o t a l : 9 + 10 + 9 + 8 + \dots + 3 = 61

Commented by naka3546 last updated on 09/Dec/21

T h a n k y o u, s i r .