Suppose once more we′re asked to choose four students from high school class of 15 to form a committee but this time we have a restriction : we don′t want to committee to consist of all seniors or all juniors .suppose there are eight seniors and seven juniors in the class. How many different committe can we form?


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Question Number 119364 by bobhans last updated on 24/Oct/20

$S u p p o s e o n c e m o r e {w e}^{'} r e a s k e d$ $t o c h o o s e f o u r s t u d e n t s f r o m h i g h s c h o o l$ $c l a s s o f 15 t o f o r m a c o m m i t t e e$ $b u t t h i s t i m e w e h a v e a r e s t r i c t i o n$ $: w e {d o n}^{'} t w a n t t o c o m m i t t e e t o$ $c o n s i s t o f a l l s e n i o r s o r a l l j u n i o r s$ $. s u p p o s e t h e r e a r e e i g h t s e n i o r s$ $a n d s e v e n j u n i o r s i n t h e c l a s s . H o w m a n y$ $d i f f e r e n t c o m m i t t e c a n w e f o r m ?$
	Answered by john santu last updated on 24/Oct/20

	$T h i s c l e a r l y c a l l f o r s u b t r a c t i o n p r i n c i p l e$ $T h e t o t a l n u m b e r o f p o s s i b l e c o m m i t t e i s$ $(\begin{matrix} 15 \\ 4 \end{matrix}) = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 1, 365$ $T h e n u m b e r o f c o m m i t t e c o n s i s t i n g a l l s e n i o r i s$ $(\begin{matrix} 8 \\ 4 \end{matrix}) = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$ $a n d t h e n u m b e r o f c o m m i t t e c o n s i s t i n g a l l$ $j u n i o r s i s (\begin{matrix} 7 \\ 4 \end{matrix}) = \frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1} = 35$ $i f w e e x l u d e t h o s e, w e s e e t h a t t h e$ $n u m b e r o f a l l o w a b l e c o m m i t t e i s$ $1, 365 - 70 - 35 = 1, 260$
	Commented by bobhans last updated on 24/Oct/20

	$n i c e . . .$
	Answered by mr W last updated on 24/Oct/20

	$C_{1}^{7} C_{3}^{8} + C_{2}^{7} C_{2}^{8} + C_{3}^{7} C_{1}^{8} = 1260$
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