The total number of numbers, lying between 100 and 1000 that can be formed with the digits 1, 2, 3, 4, 5, if the repetition of digits is not allowed and numbers are divisible by either 3 or 5, is _____________.

Solution

Maths sir · 4 steps

Count the multiples of 3 and of 5 separately, then subtract the numbers you counted twice.

1. Divisible by 3. A number is divisible by 3 when its digit sum is. Out of the 10 possible sets of three digits from 1 to 5, only four work: {1,2,3} (sum 6), {1,3,5} (sum 9), {2,3,4} (sum 9) and {3,4,5} (sum 12). Each set can be arranged in 3! = 6 ways, giving 4 × 6 = 24 numbers.
2. Divisible by 5. The units digit must be 5, since 0 is not available. The other two places take two of the remaining 4 digits in order, so there are 4 × 3 = 12 such numbers.
3. Counted twice. Numbers divisible by both 3 and 5 come from the sets that contain 5: {1,3,5} and {3,4,5}. With 5 fixed in the units place, each set gives 2 numbers: 135, 315, 345, 435. That is 4 numbers.

   n(A ∪ B) = n(A) + n(B) - n(A ∩ B) = 24 + 12 - 4

4. So. The number of such numbers is 32.

The common slip: 36. Adding 24 and 12 without removing the 4 numbers divisible by 15 gives 36, not 32.

The idea to keep. Whenever a question says "divisible by either this or that", use inclusion and exclusion. Add the two counts, then subtract the numbers that satisfy both conditions.

Answer from the NTA final answer key. Solution by PrepWiser.