How-many-numbers-divisible-by-5-can-be-made-with-the-digits-2-3-4-and-5-where-no-digit-is-being-used-more-than-once-in-each-number-

Question Number 55119 by pieroo last updated on 17/Feb/19

How many numbers, divisible by 5, can

be made with the digits 2, 3, 4 and 5 where

no digit is being used more than once

in each number ?

Commented by Tawa1 last updated on 17/Feb/19

For a number to be divisible by 5 . The number must end with 5 .

∴ for all the four digit number (2, 3, 4, 5) .

3 \times 2 \times 1 (fixed 5) = 6 ways

- - - -

∴ for three digit number (2, 3, 4, 5) .

3 \times 2 (fixed 5) = 6 ways

- - -

∴ for two digit number (2, 3, 4, 5) .

3 (fixed 5) = 3 ways

- -

∴ for one digit number (2, 3, 4, 5) .

(only 5) = 1 way

-

Therefore, Number of numbers divisible by 5 = 6 + 6 + 3 + 1

Therefore, Number of numbers divisibke by 5 = 16 numbers

Answered by math1967 last updated on 17/Feb/19

3! = 6 (4 d i g i t n u m b e r)

{\overset{3}{P}}_{2} = 6 (3 d i g i t n u m b e r s)

{\overset{3}{P}}_{1} = 3 (2 d i g i t n u m b e r s)

1 d i g i t n u m b e r = 1

t o t a l = 6 + 6 + 3 + 1 = 16

Commented by pieroo last updated on 18/Feb/19

Thanks sir

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