The number of permutations, of the digits 1, 2, 3, ..., 7 without repetition, which neither contain the string 153 nor the string 2467, is ___________.

Solution

Maths sir · 4 steps

Glue each forbidden string into one block, count the bad ones, then subtract them from 7!.

1. All arrangements. With no restriction, the 7 distinct digits can be arranged in 7! = 5040 ways.
2. Glue each string. To contain 153, treat it as one block. The block plus 2, 4, 6, 7 makes 5 units, so there are 5! = 120 arrangements. To contain 2467, treat it as one block. The block plus 1, 3, 5 makes 4 units, so there are 4! = 24 arrangements.
3. Both strings. The two strings share no digit and together use all 7 digits. So with both glued there are only 2 units, giving 2! = 2 arrangements. These were counted in both groups above, so add them back once.

   n(A ∪ B) = 120 + 24 - 2 = 142

4. So. The permutations that contain neither string number 5040 - 142 = 4898.

The idea to keep. To count arrangements containing a string, glue it into one block. For "neither", subtract the union from the total, and always add back the overlap.

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