Question

how many 3 digit even numbers are there such that if 3 is one of the digits, then 5 is the next digit

Answer 1

Answer:

365.

Explanation;

We want to find the number of 3-digit numbers formed by 10 different digits (0 to 9) satisfying the given conditions. Repetition of digits is allowed. Since we need even numbers, so the unit's place can be filled by any of the digits 0, 2, 4, 6, 8 i.e. in 5 ways.

Now if 3 is at ten's place, then as per the given condition, the digit 5 should be at unit's and in this case the number will not be an even number. So, 3 cannot be at ten's place. Therefore, 3 can only be at hundred's place.

Two cases arise:

Case I:

When 3 is at hundred's place.

if 3 is at hundred's place, then 5 must be at ten's place. Unit's place can be filled by any of digits 0, 2, 4, 6, 8 i.e. in 5 wavs.

So,The number of such even numbers = 1x 1x 5= 5.

Case II.

When 3 is not at hundred's place.

In this case, hundred's place can be filled in 8 ways (as 0 and 3 cannot be used at hundred'splace). Ten's place can be filled in 9 ways (as 3 cannot be used at this place). Unit's place can be

flled in 5 ways (any digit 0, 2, 4, 6, 8 can be used at this place).

So Number of such even numbers = 8 x 9 x 5 =360.

So,The number of required numbers 5 + 360=365.