Find the number of integers greater than 6200 that can be formed from the digits 1,3,6,8 and 9, where each digit is used at most once.


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Question Number 199333 by necx122 last updated on 01/Nov/23

$F i n d t h e n u m b e r o f i n t e g e r s g r e a t e r$ $t h a n 6200 t h a t c a n b e f o r m e d f r o m$ $t h e d i g i t s 1, 3, 6, 8 a n d 9, w h e r e e a c h$ $d i g i t i s u s e d a t m o s t o n c e .$
	Answered by aleks041103 last updated on 01/Nov/23

	$C a s e 1 : 4 - d i g i t n u m s$ $T h e f i r s t d i g i t c a n b e o n l y 6, 8 o r 9 .$ $C a s e 1.1 : 8 o r 9$ $w e h a v e 4 d i g i t s l e f t f o r 3 p o s i t i o n s .$ $\Rightarrow i n t o t a l = 2 \times (4 \times 3 \times 2) = 48$ $C a s e 1.2 : 6$ $t h e n u m b e r c a n b e o f t y p e s :$ $63 * , 68 , 69 *$ $f o r e a c h o f w h i c h w e h a v e 6 o p t i o n s$ $\Rightarrow i n t o t a l = 3 \times 6 = 18$ $C a s e 2 : 5 - d i g i t n u m s$ $A l l 5 - d i g i t a r e g r e a t e r t h n 6200 .$ $\Rightarrow i n t o t a l = 5 \times 4 \times 3 \times 2 \times 1 = 120$ $A n s w e r = 48 + 18 + 120$ $\Rightarrow A n s . = 186$
	Commented by necx122 last updated on 01/Nov/23
	Thank you. This reminds me of how I learnt this those days with Mr. W's method. Thank you so much our old friend, Aleks.
	Answered by mr W last updated on 01/Nov/23

	$1) 5 - d i g i t n u m b e r s :$ $5! = 120$ $2) v a l i d 4 - d i g i t n u m b e r s :$ $63 x y, 68 x y, 69 x y,$ $81 x y, 83 x y, 86 x y, 89 x y,$ $91 x y, 93 x y, 96 x y, 98 x y$ $x y = 2 f r o m 3 = P_{2}^{3}$ $\Rightarrow 11 \times P_{2}^{3} = 66$ $t o t a l l y : 120 + 66 = 186 ✓$
	Commented by necx122 last updated on 01/Nov/23
	always on it sir. Thank you. This is clear and well understood.
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