The integers between 1 to 10^4 contain exactly one 8 and one 9 is? I got 1160 but one is arguing 1154only..kindly help me out


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Question Number 180335 by SLVR last updated on 10/Nov/22

$T h e i n t e g e r s b e t w e e n 1 t o 10^{4}$ $c o n t a i n e x a c t l y o n e 8 a n d o n e 9$ $i s ? I g o t 1160 b u t o n e i s a r g u i n g$ $1154 o n l y . . k i n d l y h e l p m e o u t$
Commented by mr W last updated on 10/Nov/22

$i t h i n k b o t h a r e w r o n g, o r i {d i d n}^{'} t$ $u n d e r s t a n d t h e q u e s t i o n c o r r e c t l y .$
	Answered by mr W last updated on 10/Nov/22

	$1) 2 d i g i t n u m b e r s :$ $89 a n d 98 \Rightarrow 2 n u m b e r s$ $2) 3 d i g i t n u m b e r s :$ $Y 89, 8 X 9, 89 X$ $\Rightarrow 2 \times (7 + 8 + 8) = 46 n u m b e r s$ $X = 0, 1, 2, . . ., 7$ $Y = 1, 2, . . ., 7$ $3) 4 d i g i t n u m b e r s :$ $Y X 89 \Rightarrow 2 \times 7 \times 8 = 112$ $Y 8 X 9 \Rightarrow 2 \times 7 \times 8$ $Y 89 X \Rightarrow 2 \times 7 \times 8$ $8 X X 9 \Rightarrow 2 \times 8 \times 8 = 128$ $8 X 9 X \Rightarrow 2 \times 8 \times 8$ $89 X X \Rightarrow 2 \times 8 \times 8$ $\Rightarrow 3 \times 112 + 3 \times 128 = 720$ $t o t a l l y : 2 + 46 + 720 = 768$
	Commented by SLVR last updated on 11/Nov/22

	$T h a n k s . . . P r o f . W . . r e a l l y i c o u n t e d$ $r e p e a t e d l y . . i a m w r o n g s i r . . .$
	Answered by Acem last updated on 10/Nov/22

	$Y o u h a v e t h r e e g e n e r a l c a s e s$ $2 d i g 3 d i g 4 d i g$ $89 x 89 x x 89 w i t h 2! {p e r m u t a t i o n}^{8, 9}$ $x 89 3! = 3 \times 2^{8, 9}$ $x x 89 \frac{4!}{2!} = 12 = 6 \times 2^{8, 9}$ $w e l l {l e t}^{'} s n o t e c a s e s$ $∙ x^{'} 89, 8 x 9, 89 x ∣_{x : 0 \to 7}^{x^{'} : 1 \to 7}$ $∙ x^{'} x 89, x^{'} 8 x 9, x^{'} 89 x ∣_{x^{'} : 1 \to 7}^{x : 0 \to 7}$ $8 x x 9, 8 x 9 x, 89 x x^{x : 0 \to 7}$ ${N u m}_{N u m b .} = 2 [1 + (7 + 2 \times 8) + 3 (7 \times 8 + 8 \times 8)]$ $= 2 [1 + 23 + 24 (15)] = 768 n u m b e r s$
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