Using only the integers 4 to 8, how many even numbers can be formed if each must lie between 4000 and 9000?


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Question Number 27513 by NECx last updated on 08/Jan/18

$U s i n g o n l y t h e i n t e g e r s 4 t o 8,$ $h o w m a n y e v e n n u m b e r s c a n b e$ $f o r m e d i f e a c h m u s t l i e b e t w e e n$ $4000 a n d 9000 ?$
	Answered by Rasheed.Sindhi last updated on 08/Jan/18

	$A s t h e n u m b e r s a r e b e t w e e n$ $4000 & 9000, s o t h e n u m b e r o f$ $p o s i t i o n s i s 4 .$ $E a c h o f 1000 t h, 100 t h & 10 t h$ $p o s i t i o n s c a n b e f i l l e d w i t h$ $4, 5, 6, 7 & 8 i - e i n 5 w a y s b u t$ $t h e u n i t p o s i t i o n c a n b e f i l l e d$ $w i t h 4, 6 & 8 i - e i n 3 w a y s b e c a u s e$ $t h e n u m b e r s a r e e v e n .$ $∴ T o t a l w a y s = 5.5.5.3 = 375$
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