Question

the number of 3 digit multiples of 5 such that no digit is prime​

Answer 1

Answer:

This can be solved with help of permutation formula

nPr = n!/ (n-1)!

Where, n are alternatives available

And r is the number of alternatives to choose.

1st digit, n= 9 (number of alternatives available)

i.e (1,2,3,4,5,6,7,8,9 we cannot include 0 because the 3digit number will become 2digit number if we place 0 in 1st place)

r = 1 (since we only need one alternative to choose)

So, nPr = 9P1 = 9 (as per the formula)

2nd digit, n= 10 ( 0,1,2,3,4,5,6,7,8,9) r=1

So, nPr = 10P1 = 10

For a number to be multiple of 5 the last digit of the number should be 0 or 5

3rd digit, n= 2 (0,5) r=1

So, nPr = 2P1 = 2

9*10*2 = 180

Step-by-step explanation: