Between 20000 and 70000 find the number of even integers in which no digits is repeated


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Question Number 124709 by benjo_mathlover last updated on 05/Dec/20

$B e t w e e n 20000 a n d 70000$ $f i n d t h e n u m b e r o f e v e n i n t e g e r s$ $i n w h i c h n o d i g i t s i s r e p e a t e d$
	Answered by liberty last updated on 05/Dec/20
	$Let abcde be a required even integer. As shown in the following diagram, the 1^(st) digit can be chosen from { 2,3,4,5,6} and the 5^(th) digit e can be chosen from {0,2,4,6,8}. ⌊ a ⌋ ⌊ b c d ⌋ ⌊e ⌋ since {2,3,4,5,6} ∩ { 0,2,4,6,8}= {2,4,6} we devide the problem into 2 disjoint cases case(1) a ∈ {2,4,6} there are = 3×4×P _3^8 case(2) a ∈ {3,5} there are = 2×5×P _3^8 The total number of required even numbers is 4032 + 3360 = 7392 .$
	$L e t a b c d e b e a r e q u i r e d e v e n i n t e g e r .$ $A s s h o w n i n t h e f o l l o w i n g d i a g r a m, t h e$ $1^{s t} d i g i t c a n b e c h o s e n f r o m {2, 3, 4, 5, 6}$ $a n d t h e 5^{t h} d i g i t e c a n b e c h o s e n f r o m$ ${0, 2, 4, 6, 8} .$ $⌊ a ⌋ ⌊ b c d ⌋ ⌊ e ⌋$ $s i n c e {2, 3, 4, 5, 6} \cap {0, 2, 4, 6, 8} = {2, 4, 6}$ $w e d e v i d e t h e p r o b l e m i n t o 2 d i s j o i n t c a s e s$ $c a s e (1) a \in {2, 4, 6} t h e r e a r e = 3 \times 4 \times P_{3}^{8}$ $c a s e (2) a \in {3, 5} t h e r e a r e = 2 \times 5 \times P_{3}^{8}$ $T h e t o t a l n u m b e r o f r e q u i r e d e v e n n u m b e r s$ $i s 4032 + 3360 = 7392 .$
	Answered by mr W last updated on 06/Dec/20
	$ABCDE A: 2,3,4,5,6 E: 0,2,4,6,8 case 1: E∈{0,8} ⇒2×5×P_3 ^8 =3360 case 2: E∈{2,4,6} ⇒3×4×P_3 ^8 =4032 ⇒totally 3360+4032=7392$
	$A B C D E$ $A : 2, 3, 4, 5, 6$ $E : 0, 2, 4, 6, 8$ $c a s e 1 : E \in {0, 8}$ $\Rightarrow 2 \times 5 \times P_{3}^{8} = 3360$ $c a s e 2 : E \in {2, 4, 6}$ $\Rightarrow 3 \times 4 \times P_{3}^{8} = 4032$ $\Rightarrow t o t a l l y 3360 + 4032 = 7392$
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