The sum of the digits in the unit place of all four digit numbers formed using the numbers 3 4 5 andL 6 without repetition is a)432 b)108 c)36 d)18


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Question Number 52379 by Necxx last updated on 07/Jan/19

$T h e s u m o f t h e d i g i t s i n t h e u n i t$ $p l a c e o f a l l f o u r d i g i t n u m b e r s$ $f o r m e d u s i n g t h e n u m b e r s 3 4 5 a n d L$ $6 w i t h o u t r e p e t i t i o n i s$ $a) 4 32 b) 108 c) 36 d) 18$
	Answered by tanmay.chaudhury50@gmail.com last updated on 07/Jan/19

	$4 o u r d i g i t n u m b e r s = 4! = 24$ $w h i l e a d d i n g w e h a v e t o a d d e a c h d i g i t (3, 4, 5, 6)$ $6 t i m e s$ $s o s u m o f d i g u t i n u n i t p l a c e i s$ $6 (3 + 4 + 5 + 6) = 108$
	Commented by Necxx last updated on 07/Jan/19

	$s o r r y f o r m y f u n n y q u e s t i o n s b u t$ $w h y d i d y o u m u l t i p l y b y 6 t i m e s .$ $W h e n I s o l v e d I h a d 18 b y j u s t$ $a d d i n g a l l t h e d i g i t s b u t y o u r$ $a n s w e r i s t h e c o r r e c t o n e . T h a n k s$ $i n a d v a n c e .$
	Commented by mr W last updated on 07/Jan/19

	$t h e r e a r e 3! = 6 n u m b e r s w h o s e l a s t$ $d i g i t i s 3, 4, 5, 6 r e s p e c t i v e l y .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 08/Jan/19

	$- - - \overset{3}{-} u n i t p l a c e 3 \to (3 \times 2 \times 1) = 6 n o s$ $- - - \overset{4}{-} u n i t p l a c e 4 \to (3 \times 2 \times 1) = 6 n o s$ $- - - \overset{5}{-} u n i t p l a c e 5 \to (3 \times 2 \times 1) = 6 n o s$ $- - - \overset{6}{-} u n i t p l a c e 6 \to (3 \times 2 \times 1) = 6 n o s$ $n o w a d d t h e m . . . . y o u h a v e t o a d d = 6 (3 + 4 + 5 + 6) = 108 . .$
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