2n objects of each of three kinds are given to two persons, so that each person gets 3n objects. Prove that this can be done in 3n^2 + 3n + 1 ways.


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Question Number 170794 by mr W last updated on 30/May/22

$2 n objects of each of three kinds are$ $given to two persons, so that each$ $person gets 3 n objects . Prove that$ $this can be done in 3 n^{2} + 3 n + 1 ways .$
	Answered by aleks041103 last updated on 05/Jun/22
	$let the first person have a,b and c objects from the 1st, 2nd and 3rd kind ⇒a+b+c=3n 0≤a,b,c≤2n b+c=3n−a 1) for a=0,...,n (b,c)∈{(n−a,2n),...,(2n,n−a)} ⇒there are 2n−(n−a)+1=n+a+1 ways for given a ⇒ for all a=0,...,n we have in total: Σ_(a=0) ^n (n+1+a)=(n+1)^2 +((n(n+1))/2) 2) for a=n+1,...,2n (b,c)∈{(0,3n−a),...,(3n−a,0)} ⇒ there are 3n+1−a ways for given a ⇒for all a=n+1,...,2n, in tot.: Σ_(a=n+1) ^(2n) (3n+1−a)=Σ_(k=1) ^n (3n+1−(k+n))= =Σ_(k=1) ^n (2n+1−k)=(2n+1)(n)−((n(n+1))/2) 3) total: ways=(n+1)^2 +((n(n+1))/2)+(2n+1)n−((n(n+1))/2)= =(n+1)^2 +2n^2 +n= =n^2 +2n+1+2n^2 +n= =3n^2 +3n+1$
	$l e t t h e f i r s t p e r s o n h a v e a, b a n d c o b j e c t s$ $f r o m t h e 1 s t, 2 n d a n d 3 r d k i n d$ $\Rightarrow a + b + c = 3 n$ $0 ⩽ a, b, c ⩽ 2 n$ $b + c = 3 n - a$ $1) f o r a = 0, . . ., n$ $(b, c) \in {(n - a, 2 n), . . ., (2 n, n - a)}$ $\Rightarrow t h e r e a r e 2 n - (n - a) + 1 = n + a + 1 w a y s$ $f o r g i v e n a$ $\Rightarrow f o r a l l a = 0, . . ., n w e h a v e i n t o t a l :$ $\underset{a = 0}{\sum^{n}} (n + 1 + a) = {(n + 1)}^{2} + \frac{n (n + 1)}{2}$ $2) f o r a = n + 1, . . ., 2 n$ $(b, c) \in {(0, 3 n - a), . . ., (3 n - a, 0)}$ $\Rightarrow t h e r e a r e 3 n + 1 - a w a y s f o r g i v e n a$ $\Rightarrow f o r a l l a = n + 1, . . ., 2 n, i n t o t . :$ $\underset{a = n + 1}{\sum^{2 n}} (3 n + 1 - a) = \underset{k = 1}{\sum^{n}} (3 n + 1 - (k + n)) =$ $= \underset{k = 1}{\sum^{n}} (2 n + 1 - k) = (2 n + 1) (n) - \frac{n (n + 1)}{2}$ $3) t o t a l :$ $w a y s = {(n + 1)}^{2} + \frac{n (n + 1)}{2} + (2 n + 1) n - \frac{n (n + 1)}{2} =$ $= {(n + 1)}^{2} + 2 n^{2} + n =$ $= n^{2} + 2 n + 1 + 2 n^{2} + n =$ $= 3 n^{2} + 3 n + 1$
	Commented by mr W last updated on 05/Jun/22

	$t h a n k s s i r!$
	Answered by mr W last updated on 06/Jun/22

	${(1 + x + x^{2} + x^{3} + . . . + x^{2 n})}^{3}$ $= \frac{{(1 - x^{2 n + 1})}^{3}}{{(1 - x)}^{3}}$ $= (1 - 3 x^{2 n + 1} + 3 x^{4 n + 2} - x^{6 n + 3}) \underset{k = 0}{\sum^{\infty}} C_{2}^{k + 2} x^{k}$ $c o e f f i c i e n t o f x^{3 n} t e r m :$ $C_{2}^{3 n + 2} - 3 C_{2}^{n + 1}$ $= \frac{(3 n + 2) (3 n + 1) - 3 (n + 1) (n)}{2}$ $= \frac{9 n^{2} + 9 n + 2 - 3 n^{2} - 3 n}{2}$ $= 3 n^{2} + 3 n + 1$
	Commented by aleks041103 last updated on 05/Jun/22

	$T h i s i s v e r y c l e v e r! T h a n k y o u!$
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