Question-57044

Question Number 57044 by mr W last updated on 29/Mar/19

Commented by mr W last updated on 29/Mar/19

r e p o s t e d

Q u e s t i o n f r o m T i n k u t a r a s i r

Answered by tanmay.chaudhury50@gmail.com last updated on 29/Mar/19

n u m b e r o f A t y p e o b j e c t = 2 n

n u m b s e r o f B t y p e o b j e c t = 2 n

n u m b e r o f C t y p e o b j e c t = 2 n

s o t o t a l o b j e c t = 6 n

t o b e d i s t r i b u t e d b e t w e e n t w o p e r s o n .

(1 + x + x^{2} + . . + x^{2 n}) (1 + x + x^{2} + \dots + x^{2 n}) (1 + x + x^{2} + . . x^{2 n})

\frac{1 - x^{2 n + 1}}{1 - x} \times \frac{1 - x^{2 n + 1}}{1 - x} \times \frac{1 - x^{2 n + 1}}{1 - x}

= \frac{{(1 - x^{2 n + 1})}^{3}}{{(1 - x)}^{3}}

= {(1 - x)}^{- 3} {(1 - x^{2 n + 1})}^{3}

r e q u i r e d a n s w e r i s t h e c o e f f i c i e n t o f

x^{3 n} i n {(1 - x)}^{- 3} {(1 - x^{2 n + 1})}^{3}

= (1 + 3 x + 6 x^{2} + . . + \frac{(r + 1) (r + 2)}{1 \times 2} x^{r} + . .) (1 - 3 x^{2 n + 1} + 3 x^{4 n + 2} - x^{6 n + 3})

= \frac{(3 n + 1) (3 n + 2)}{1 \times 2} x^{3 n} - 3 x^{2 n + 1} \times \frac{(n - 1 + 1) (n - 1 + 2)}{1 \times 2} x^{n - 1}

= x^{3 n} {\frac{9 n^{2} + 9 n + 2}{2} - \frac{3}{2} \times (n^{2} + n)}

= \frac{x^{3 n}}{2} (9 n^{2} + 9 n + 2 - 3 n^{2} - 3 n)

= \frac{x^{3 n}}{2} (6 n^{2} + 6 n + 2)

s o a n s w e r i s 3 n^{2} + 3 n + 1

Commented by mr W last updated on 30/Mar/19

t h a n k s s i r! n i c e p r o o f!

\frac{{(1 - x^{2 n + 1})}^{3}}{{(1 - x)}^{3}}

= (1 - 3 x^{2 n + 1} + 3 x^{4 n + 2} - x^{6 n + 3}) \underset{k = 0}{\sum^{\infty}} C_{k}^{2 + k} x^{k}

x^{3 n} : C_{3 n}^{2 + 3 n} - 3 C_{n - 1}^{n + 1} = C_{2}^{3 n + 2} - 3 C_{2}^{n + 1}

= \frac{(3 n + 2) (3 n + 1) - 3 (n + 1) n}{2}

= \frac{6 n^{2} + 6 n + 2}{2}

= 3 n^{2} + 3 n + 1

Commented by tanmay.chaudhury50@gmail.com last updated on 30/Mar/19

t h i s p l a t f o r m b i n d u s b y i n v i s b l e b o n d a n d b o o s t

u s t o q u e n c h o u r t h i r s t o f q u e s t t o f i n d t h e t r u t h \dots

t h e s o l u t i o n \dots