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Question Number 184478 by Shrinava last updated on 07/Jan/23

	Answered by mr W last updated on 08/Jan/23

	$‘ ‘ g e n e r a t i n g {f u n c t i o n}^{″} m e t h o d$ ${(x + x^{2} + x^{3} + . . .)}^{4} = \frac{x^{4}}{{(1 - x)}^{4}} = x^{4} \underset{k = 0}{\sum^{\infty}} C_{3}^{k + 3} x^{k}$ $4 + k = 26 \Rightarrow k = 22$ $C_{3}^{22 + 3} = \frac{25 \times 24 \times 23}{6} = 2300 ✓$ $o r u s i n g ‘ ‘ b a r s & {s t a r s}^{″} m e t h o d :$ $C_{r - 1}^{n - 1} = C_{4 - 1}^{26 - 1} = C_{3}^{25} = 2300 ✓$
	Commented by Shrinava last updated on 07/Jan/23

	$thank you dear professor$ $r = number of roots ? please$
	Commented by mr W last updated on 07/Jan/23

	$t o p u t n b a l l s i n t o r b o x e s t h e r e a r e$ $C_{r - 1}^{n - 1} w a y s .$
	Commented by Shrinava last updated on 07/Jan/23

	$dear professor,$ $stars and bars theorem :$ $x_{1} + x_{2} + . . . x_{k} = r \Rightarrow (\begin{matrix} r + k - 1 \\ k - 1 \end{matrix})$ $\Rightarrow (\begin{matrix} 26 + 4 - 1 \\ 4 - 1 \end{matrix}) = (\begin{matrix} 29 \\ 3 \end{matrix}) = \frac{29!}{3! \cdot 26!} = 3654 ?$
	Commented by mr W last updated on 08/Jan/23

	$e x a m p l e :$ $x_{1} + x_{2} + x_{3} = 8$ $x_{i} ⩾ 1$ $i t m e a n s i n h o w m a n y w a y s c a n 8$ $a p p l e s b e d i s t r i b u t e d t o 3 p e r s o n s$ $s u c h t h a t e a c h p e r s o n g e t s a t l e a s t$ $o n e a p p l e ?$ $i n f o l l o w i n g a s t a r ★ m e a n s a n a p p l e .$ $t o d i s t r i b u t e 8 a p p l e s i n t o 3 p o r t i o n s$ $w e j u s t n e e d t o s e p a r a t e t h e 8 s t a r s$ $w i t h 2 b a r s, l i k e t h i s :$ $★ ∣ ★ ★ ★ ★ ★ ∣ ★ ★$ $w e s e e t h e r e a r e 7 p o s i t i o n s w h e r e w e$ $c a n p l a c e t h e 2 b a r s, t h e r e f o r e t h e r e$ $a r e C_{2}^{7} w a y s t o d i s t r i b u t e 8 a p p l e s t o$ $3 p e r s o n s . g e n e r a l l y t h e r e a r e C_{r - 1}^{n - 1}$ $w a y s t o d i s t r i b u t e n i d e n t i c a l o b j e c t s$ $i n t o r d i f f e r e n t b o x e s s u c h t h a t e a c h$ $b o x g e t s a t l e a s t o n e o b j e c t .$ $t h i s m e t h o d i s c a l l e d ‘ ‘ s t a r s & {b a r s}^{″}$ $m e t h o d .$ $s o x_{1} + x_{2} + . . . x_{r} = n w i t h x_{i} \in N$ $h a s C_{r - 1}^{n - 1} = (\begin{matrix} n - 1 \\ r - 1 \end{matrix}) s o l u t i o n s .$
	Commented by mr W last updated on 08/Jan/23

	$i h a v e u s e d t w o m e t h o d s :$ $‘ ‘ g e n e r a t i n g {f u n c t i o n}^{″} m e t h o d a n d$ $‘ ‘ s t a r s & {b a r s}^{″} m e t h o d .$ $t h e ‘ ‘ s t a r s & {b a r s}^{″} m e t h o d i s$ $e x p l a i n e d a b o v e . {i t}^{'} s e a s y t o$ $u n d e r s t a n d . b u t y o u c a n u s e$ $i t o n l y f o r s o l v i n g e a s y c a s e s .$ $t h e g e n e r a t i n g f u n c t i o n m e t h o d i s$ $m u c h m o r e p o w e r f u l, a n d c a n s o l v e$ $c o m p l i c a t e d c a s e s l i k e t h i s :$ $x_{1} + x_{2} + x_{3} + x_{4} = 26$ $a l l n u m b e r s a r e n a t u r a l n u m b e r s,$ $b u t x_{1} a r e o d d, x_{2} a r e e v e n e q u a l o r$ $g r e a t e r t h a n 6, w h i l e x_{4} a r e a t m o s t 5 .$ $t h e n u m b e r o f s o l u t i o n s u n d e r t h e s e$ $c o n d i t i o n s i s t h e c o e f f i c i e n t o f t e r m$ $x^{26} i n t h e e x p a n s i o n o f f o l l o w i n g$ $e x p r e s s i o n :$ $(x + x^{3} + x^{5} + . . .) (x^{6} + x^{8} + x^{10} + . . .) (x + x^{2} + x^{3} + . . .) (x + x^{2} + x^{3} + x^{4} + x^{5})$ $= \frac{x}{1 - x^{2}} \times \frac{x^{6}}{1 - x^{2}} \times \frac{x}{1 - x} \times \frac{x (1 - x^{5})}{1 - x}$ $= \frac{x^{9} (1 - x^{5})}{{(1 - x)}^{2} {(1 - x^{2})}^{2}}$ $t h e c o e f f i c i e n t o f x^{26} i s 190 . t h a t$ $m e a n s t h e r e a r e 190 s o l u t i o n s .$
	Commented by Shrinava last updated on 08/Jan/23

	$cool dear professor thank you$
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