If ^((a^2 −a)) C_2 =^((a^2 −a)) C


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Question Number 97737 by bagjamath last updated on 09/Jun/20

$If^{(a^{2} - a)} C_{2} =^{(a^{2} - a)} C_{4}, then a =$
Commented by som(math1967) last updated on 09/Jun/20

$a = 3$
Commented by smridha last updated on 09/Jun/20

$y o u h a v e t o p u t . . i n w h i c h s e t$ $d o e s^{'} a^{'} b e l o n g ? ?$
Commented by mr W last updated on 09/Jun/20

$a^{2} - a = 2 + 4 = 6$ $(a + 2) (a - 3) = 0$ $\Rightarrow a = 3$
	Answered by smridha last updated on 09/Jun/20

	$\frac{(a^{2} - a)!}{2! (a^{2} - a - 2)!} = \frac{(a^{2} - a)!}{4! (a^{2} - a - 4)!}$ $(a^{2} - a - 2) (a^{2} - a - 3) = 12$ $l e t (a^{2} - a - 2) = x$ $x^{2} - x - 12 = 0 x = \frac{1 \underline{+} 7}{2} x = (4, - 3)$ $f o r x = 4$ $a^{2} - a - 6 = 0 a = \frac{1 \underline{+} 5}{2} s o a = (3, - 2)$ $f o r x = - 3$ $a^{2} - a + 1 = 0 s o a = \frac{1 \underline{+} i \sqrt{3}}{2} a = (e^{\frac{i π}{3}}, e^{\frac{- i π}{3}})$
	Commented by smridha last updated on 09/Jun/20

	$I a m n o t f i g h t i n g w i t h o t h e r s . . .$ $b u t y o u t r i e d t o d o s o I t h i n k . .$ $I j u s t w a n t e d t o t e l l t w o s i m p l e$ $t h i n g s (i) I a m n o t c l a i m a l l t h e$ $p o s s i b l e v a l u e s o f a a r e t h e$ $s o l u t i o n s (i i) e v e n I d i d n o t c o n c l u d e$ $a n y t h i n g s o w h y y o u a r e t h e g u y s$ $a r e s o w o r r i d w i t h t h a t . . .$ $a m I c r i t i c i z e a n y o n e ? ? i f t h i s$ $i s s o s i m p l e m a t t e r t h e n i t i s n o t$ $r e q u a r e f u r t h u r c o m m e n t .$
	Commented by mathmax by abdo last updated on 09/Jun/20

	$how to get complex number in combination . . .!$
	Commented by smridha last updated on 09/Jun/20

	$a m I p u t c o m p l e x n u m b e r i n$ $c o m b i n a t i o n ? ? ? t h e n w h a t ? ?$ $I j u s t s h o w h o w m a n y v a l u e s$ $a h a v e ? ? I {d o n}^{'} t c l a i m a l l t h e$ $p o s s i b l e v a l u e s m u s t s a t i s f y$ $t h e c o m b i n a t i o n . . . . . .$ $t h e a p p e r e n c e o f t h i s q u e s t i o n i s$ $w r o n g!!! f i n d a ? ? ? w h a t d o e s i t$ $m e a n ? ? w e l o g i c a l l y c h o o s e$ $t h e v a l u e o f a a m o n g t h e v a l u e s$ $o f a . s o i t d o e s n o t m e a n t h a t a$ $h a v e o n l y p o s s i b l e v a l u e 3 . .$ $m o r e o v e r i t m e a n s a t a k e s v a l u e$ $3 f o r s a t i s f y t h i s c o m b i n a t i o n .$ $f o r t h i s c o m b i n a t i o n 3 i s u n i q u e$ $v a l u e (o f a) b u t n o t f o r a .$
	Commented by mr W last updated on 09/Jun/20

	$a s f o r t h e a n s w e r o f t h i s q u e s t i o n,$ $t h e o n l y c o r r e c t a n s w e r i s a = 3, o t h e r$ $v a l u e s o f a a r e n o n - s e n s e f o r t h i s$ $q u e s t i o n .$ $i t h i n k a s C_{2}^{a^{2} - a} = C_{4}^{a^{2} - a} i s w r i t t e n, i t i s$ $a u t o m a t i c a l l y s a i d d u e t o d e f i n i t i o n$ $t h a t a \in Z a n d a^{2} - a ⩾ 2 a n d a^{2} - a ⩾ 4 .$
	Commented by smridha last updated on 09/Jun/20

	$o o h m a n!! a m i c l a i m a l l t h e p o s s i b l e$ $v a l u e s s a t i s f y t h i s c o m b i n a t i o n ? ?$ $e v e n a m I p u t a n y c o n c l u t i o n i n m y$ $s o l u t i o n ? ?$ $t h i s i s t h e q u e s t i o n o f h o w m u c h$ $w e c a n d o ? ?$ $w e h a v e a n y i d e a ? ? a b o u t$ $t h e f a c t o r i a l o f c o m p l e x n u m b e r!!$ $f a c t o r i a l o f n e g e t i v e i n t i g e r!!$ $s o w e i g n o r e t h e m . . . . t h i s i s t h e$ $f a u l t . i f t h e r e i s t h e o r y b e h i n d$ $t h e m w e a r e t h e n s a t i s f i e d . . i s n o t$ $i t ? ? ? ?$
	Commented by mr W last updated on 09/Jun/20

	$s i r : n o b o d y s a i d t h a t y o u a r e w r o n g$ $o r s o m e t h i n g l i k e t h a t . i t s e e m s$ $y o u s e e c r i t i c a l c o m m e n t s f r o m o t h e r s$ $a s a t t a c k s a n d w a n t t o f i g h t b a c k .$ $w h a t a p i t y! s i n c e a f o r u m i s p l a c e f o r$ $d i s c u s s i o n, e x c h a n g e a n d l e a r n i n g$ $f r o m e a c h o t h e r . a l l p e o p l e h e r e a r e$ $e q u a l .$
	Commented by abdomathmax last updated on 09/Jun/20

	$your answer is not correct$
	Commented by smridha last updated on 09/Jun/20

	$M r A b d o m a t h m a x : a m I w r i t e a n y v a l u e s a s a n s ? ?$ $i f y o u t h i n k t h a t t h i s i s y o u r$ $f u l t n o t m i n e . .!!$
	Commented by mathmax by abdo last updated on 09/Jun/20

	$dont worry and be happy we are here to help and learn . . . maths is a sea$ $without borders . . . .!$
	Answered by mathmax by abdo last updated on 09/Jun/20

	$C_{a^{2} - a}^{2} = C_{a^{2} - a}^{4} \Rightarrow \frac{(a^{2} - a)!}{2! (a^{2} - a - 2)!} = \frac{(a^{2} - a)!}{4! (a^{2} - a - 4)!} (a ⩾ 2) \Rightarrow$ $2! (a^{2} - a - 2)! = 4! (a^{2} - a - 4)! \Rightarrow$ $2 (a^{2} - a - 2) (a^{2} - a - 3) = 4! \Rightarrow (a^{2} - a - 2) (a^{2} - a - 3) = \frac{4.3.2}{2} = 12$ $let a^{2} - a - 2 = 2 u and a^{2} - a - 3 = 2 v + 1 \Rightarrow 2 v + 1 = 2 u - 1 \Rightarrow$ $2 u . (2 v + 1) = 12 \Rightarrow u (2 u - 1) = 6 \Rightarrow {2 u}^{2} - u - 6 = 0 (u integr natural)$ $Δ = 1 - 4 \times 2 \times (- 6) = 1 + 48 = 49 \Rightarrow u_{1} = \frac{1 + 7}{4} = 2$ $u_{2} = \frac{1 - 7}{4} = - \frac{3}{2} (to elminate) \Rightarrow u = 2 \Rightarrow a^{2} - a = 2 + 4 = 6 \Rightarrow a^{2} - a - 6 = 0$ $Δ = 1 - 4 (- 6) = 25 \Rightarrow a = \frac{1 + 5}{2} = 3 or a = \frac{1 - 5}{2} = - 2 < 0 (to eliminate)$ $the unique value is a = 3$
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