Permutations & combinations problem.?

You can use the principle of inclusion and exclusion.

First add all the numbers in that range that are divisible by 10. Obviously every 10th number (10, 20, 30, ...) is divisible by 10.
100 / 10 = 10 numbers

Next add all the numbers in that range that are divisible by 7. Obviously every 7th number (7, 14, 21, ...) is divisible by 7.
100 / 7 = 14 2/7 --> round down to 14 numbers

So it would seem you have 10 + 14 = 24 numbers

But wait! You accidentally double-counted at least one number. If a number is divisible by 10 *and* 7, then it was counted in both lists. Specifically the number 70 is in both lists.
Mathematically:
100 / 70 = 1 3/7 --> round down to 1.

Subtract that number that was double-counted.
10 + 14 - 1 = 23 numbers

Answer:
23 numbers are divisible by 10 or 7 in that range.

Specifically:
{7, 10, 14, 20, 21, 28, 30, 35, 40, 42, 49, 50, 56, 60, 63, 70, 77, 80, 84, 90, 91, 98, 100}

Numbers between 1 and 100 (inclusive) which are divisible by 10 are:
10, 20, 30 …….. ,70, ……., 100
No. of numbers divisible by 10 = [(100 - 10)/10] + 1 = 10

Number between 1 and 100 (inclusive) which are divisible by 7 are:
7, 14, 21, ……, 70, ……, 98
No. of numbers divisible by 7 = [(98 - 7)/7] + 1 = 14

7 × 10 = 70
70 is divisible by both 10 and 7. Hence, it is counted twice.

Total no. of numbers between 1 and 100 (inclusive) divisible by 10 or 7
= 10 + 14 - 1
= 23

Ans. 23 numbers, which are
7,10,14,20,21,28,30,35,40,42,49,50,
56,60,63,70,77,80,84,90,91,98,100.
These numbers are generated
by inserting a suitable integral
values of k1 & k2 in
1=<7k1<=100 &
1=<10k2<=100.

If a number is divisible by k, then the number is equal to nk where n is an integer.

If N is divisible by 10, then N = 10k
1 ≤ 10k ≤ 100
1/10 ≤ k ≤ 10 → 10 integers, 1, 2, ..., 10, satisfy this inequality

If N is divisible by 7, then N = 7k
1 ≤ 7k ≤ 100
1/7 ≤ k ≤ 14 2/17 → 14 integers, 1, 2, ..., 13, 14, satisfy this inequality

A total of 23 numbers between 1 and 100 inclusive are divisible either by 7 or by 10.

(The answer is not 24 because the number 70, is divisible by both 7 and 10, so the sum of 10+14 = 24 counts the number 70 twice)

Ans: 23

n(A or B) = n(A) + n(B) - n(A and B) 

now, n(divisible by 10) = 10

also, n(divisible by 7) = 14

and, n(divisible by 10 and 7) = 1

Hence, 10 + 14 - 1 = 23

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