How many permutation can be made using the word CANADA each of the six letter being used mixed in each permutation? In how many of these permutation will the three A′s be together? In how many will two A′s be together but not the three?


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Question Number 40284 by scientist last updated on 18/Jul/18

$H o w m a n y p e r m u t a t i o n c a n b e m a d e u s i n g t h e w o r d$ $C A N A D A e a c h o f t h e s i x l e t t e r b e i n g u s e d m i x e d$ $i n e a c h p e r m u t a t i o n ? I n h o w m a n y o f t h e s e$ $p e r m u t a t i o n w i l l t h e t h r e e A^{'} s b e t o g e t h e r ? I n$ $h o w m a n y w i l l t w o A^{'} s b e t o g e t h e r b u t n o t t h e t h r e e ?$
	Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jul/18

	$i) C A N A D A A = 3 C = 1 D = 1 N = 1$ $p e r m u t a i o n n u m b e r s = \frac{6!}{3!} = \frac{6 \times 5 \times 4 \times 3!}{3!} = 120$ $i i) l e t 03 n u m b e r s o f A f a v i q u i c k e d w i t h e a c h$ $o t h e r s o (A A A) C D N$ $p e r m u t a t i o n = 4! = 24$ $i i i) (A A) A C D N$ $p d r m u t a t i o n = 3 C_{2} \times 5! = 3 \times 120 = 360$ $p l s c h e c k . . .$ $a f t e r e d i t i n g . . .$ $i i i) r e q u i r e d a n s w e r i s 360 - 24 = 336$ $b e c o s o u t o f t h e s e 360 p e r m u t a t i o n 24$ $p e r m u t a t i o n a r e i n c l u d e d w h e r e 03 A A A w i l l$ $b e t o g e t h e r . . . s o a n s i s 360 - 24 = 336$
	Commented by MJS last updated on 18/Jul/18

	$i i i) might be wrong because A A A is not$ $allowed but I^{'} m not sure in these things$
	Commented by tanmay.chaudhury50@gmail.com last updated on 19/Jul/18

	$y e s y o u a r e r i g h t . . . i a m e d i t i n g . . .$
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