How-many-six-digit-numbers-contain-exactly-three-different-digits-

Question Number 124400 by mr W last updated on 03/Dec/20

H o w m a n y s i x - d i g i t n u m b e r s c o n t a i n

e x a c t l y t h r e e d i f f e r e n t d i g i t s ?

Answered by benjo_mathlover last updated on 03/Dec/20

X X X W Y Z = 9 \times C_{3}^{9} \times \frac{6!}{3!}

= 9 \times \frac{9 \times 8 \times 7}{3 \times 2 \times 1} \times 6 \times 5 \times 4 = 90720

w h e r e X \neq 0

Commented by benjo_mathlover last updated on 03/Dec/20

w h a t w r o n g s i r ?

i n p a r t X X X Y Y Z ?

Answered by mr W last updated on 03/Dec/20

t o s e l e c t t h r e e d i g i t s t h e r e a r e

C_{3}^{10} = 120 w a y s .

s a y t h e t h r e e d i g i t s a r e x, y, z .

t o f o r m a 6 d i g i t n u m b e r w i t h t h e s e

t h r e e d i g i t s w e h a v e

c a s e 1 : 4 x + y + z

\Rightarrow 3 \times \frac{6!}{4!} = 90

c a s e 2 : 3 x + 2 y + z

\Rightarrow 3 \times 2 \times \frac{6!}{3! 2!} = 360

c a s e 3 : 2 x + 2 y + 2 z

\Rightarrow \frac{6!}{2! 2! 2!} = 90

\Rightarrow 120 \times (90 + 360 + 90) = 64800

b u t i n t h e s e n u m b e r s s o m e b e g i n

w i t h z e r o :

0 x x x x y \Rightarrow 2 \times \frac{5!}{4!} = 10

0 x x x y y \Rightarrow 2 \times \frac{5!}{3! 2!} = 20

0 x x x y 0 \Rightarrow 2 \times \frac{5!}{3!} = 40

0 x x y y 0 \Rightarrow \frac{5!}{2! 2!} = 30

0 x x y 00 \Rightarrow 2 \times \frac{5!}{2! 2!} = 60

0 x y 000 \Rightarrow \frac{5!}{3!} = 20

t o s e l e c t t h e t w o d i g i t s x a n d y

t h e r e a r e C_{2}^{9} = 36 w a y s .

\Rightarrow 36 \times (10 + 20 + 40 + 30 + 60 + 20) = 6480

t o t a l v a l i d 6 d i g i t n u m b e r s :

64800 - 6480 = 58320

Commented by mr W last updated on 04/Dec/20

t h e r e i s a l s o a f o r m u l a :

\frac{9}{10} \times C_{m}^{10} \times m! \times {_{m}^{n}}

w i t h n = 6 a n d m = 3 :

\frac{9}{10} \times C_{3}^{10} \times 3! \times {_{3}^{6}} = \frac{9}{10} \times 120 \times 6 \times 90

= 58320

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