Calculate the number of ways in which (a) 5 children can be divided into groups of 2 and 3 , (b) 9 children can be divided into groups of 5 and 4, Hence calculate the number of ways in which 9 children can be divided into groups of 2,3 and 4.


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Question Number 5254 by Rasheed Soomro last updated on 02/May/16

$Calculate the number of ways in which$ $(a) 5 children can be divided into groups$ $of 2 and 3,$ $(b) 9 children can be divided into groups$ $of 5 and 4,$ $Hence calculate the number of ways in$ $which 9 children can be divided into$ $groups of 2, 3 and 4 .$
Commented by prakash jain last updated on 02/May/16

$Dividing 5 children is groups of 2 and 3 .$ $=^{5} C_{2} =^{5} C_{3}$ $Dividing 9 children is groups of 4 and 5 .$ $=^{9} C_{4} =^{9} C_{5}$ $Dividing 9 children in group of 4 and 5 .$ $=^{9} C_{5} \times^{5} C_{2}$
	Answered by Yozzii last updated on 02/May/16

	$(a) no . of ways = (\begin{matrix} 5 \\ 2 \end{matrix}) = (\begin{matrix} 5 \\ 3 \end{matrix})$ $T h i s i s b e c a u s e b y c r e a t i n g a g r o u p o f$ $2 c h i l d r e n y o u s i m u l t a n e o u s l y$ $y i e l d a s e c o n d g r o u p o f 3 c h i l d r e n$ $c o r r e s p o n d i n g t o t h e g r o u p o f t w o c h i l d r e n c h o s e n .$ $T h e a n s w e r o f (\begin{matrix} 5 \\ 3 \end{matrix}) s h o w s t h e c o n v e r s e$ $w h e r e 3 c h i l d r e n a r e c h o s e n .$ $(b) no . of ways = (\begin{matrix} 9 \\ 4 \end{matrix}) = (\begin{matrix} 9 \\ 5 \end{matrix})$ $T o d i v i d e 9 c h i l d r e n i n t o g r o u p s o f$ $2, 3 a n d 4 y o u c a n f i r s t c h o o s e$ $4 o u t o f 9 \Rightarrow (\begin{matrix} 9 \\ 4 \end{matrix}) w a y s . T h i s l e a v e s a$ $g r o u p o f 4 a n d a g r o u p o f 5 . T h e n, t o$ $s p l i t 5 i n t o g r o u p s o f 2 a n d 3 \Rightarrow (\begin{matrix} 5 \\ 3 \end{matrix}) w a y s .$ $B y t h e A N D p r i n c i p l e o f c o u n t i n g$ $w e m u s t h a v e t h e e v e n t o f c h o o s i n g$ $4 c h i l d r e n a n d t h e e v e n t o f s p l i t i n g$ $5 c h i l d r e n i n t o g r o u p s o f 2 a n d 3$ $a t t h e s a m e t i m e, i n (\begin{matrix} 9 \\ 4 \end{matrix}) \times (\begin{matrix} 5 \\ 3 \end{matrix}) = 1260 w a y s .$
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