ABCD is four digits integers . How many ABCD that suitable with A+B+C+D = 25 ?


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Question Number 57594 by naka3546 last updated on 08/Apr/19

$A B C D i s f o u r d i g i t s i n t e g e r s .$ $H o w m a n y A B C D t h a t s u i t a b l e w i t h A + B + C + D = 25 ?$
	Answered by mr W last updated on 08/Apr/19

	$A : 1, 2, . ., 9$ $B, C, D : 0, 1, 2, . ., 9$ $A + B + C + D = 25$ $(x + x^{2} + x^{3} + . . . + x^{9}) {(1 + x + x^{2} + x^{3} + . . . + x^{9})}^{3}$ $= \frac{x (1 - x^{9}) {(1 - x^{10})}^{3}}{{(1 - x)}^{4}}$ $= \frac{x (1 - x^{9}) (1 - 3 x^{10} + 3 x^{20} - x^{30})}{{(1 - x)}^{4}}$ $= \frac{x (1 - x^{9} - 3 x^{10} + 3 x^{19} + 3 x^{20} + . . .)}{{(1 - x)}^{4}}$ $= x (1 - x^{9} - 3 x^{10} + 3 x^{19} + 3 x^{20} + . . .) \underset{k = 0}{\sum^{\infty}} C_{3}^{k + 3} x^{k}$ $c o e f . o f x^{25} i s$ $C_{3}^{27} - C_{3}^{18} - 3 C_{3}^{17} + 3 C_{3}^{8} + 3 C_{3}^{7} = 2925 - 816 - 3 \times 680 + 3 \times 56 + 3 \times 35 = 342$ $\Rightarrow t h e r e a r e 342 s u i t a b l e n u m b e r s .$
	Commented by mr W last updated on 08/Apr/19

	$f o r m o r e d e t a i l s s e e Q 21800 .$
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