Q.Find how many 4 digit numbers with distinct digits can be formed using the digits 1, 2, 3, 4, 5, 7 and 9 if each number contains two even digits.

Permutations with Restrictions – A First Look

Imagine you have five friends and you want to seat them in a row of five chairs. That’s easy: $5! = 120$ ways. But now suppose two of them, say A and B, refuse to sit next to each other. Suddenly the problem feels different — you can’t just count all arrangements; you have to avoid some.

That’s the heart of permutations with restrictions: you still count arrangements of distinct objects, but certain arrangements are forbidden. The trick is to count the total and then subtract the forbidden ones, or to count directly by placing the restricted items first.

The core idea: total minus unwanted

The most powerful tool for restrictions is the complement method:

Count everything, then subtract what you don’t want.

For the "A and B never together" problem:

1. Total arrangements of 5 people: $5! = 120$
2. Count arrangements where A and B are together. Treat them as a single block: that block + the other 3 people = 4 items, arranged in $4!$ ways. Inside the block, A and B can swap: $2!$ ways. So together: $4! \times 2! = 24 \times 2 = 48$
3. Subtract: $120 - 48 = 72$

Another common restriction: fixed positions

Suppose you have 7 books and a specific book must be placed in the middle (position 4). Then:

• Fix that book in position 4: 1 way
• Arrange the remaining 6 books in the other 6 positions: $6!$ ways

Answer: $6! = 720$

If instead a book cannot be in the middle, count total ($7!$) minus arrangements where it is in the middle ($6!$): $7! - 6! = 5040 - 720 = 4320$

When restrictions involve multiple conditions

Sometimes you have two restrictions at once. For example: arrange the letters of the word MATHS such that vowels (A) are never together and consonants (M, T, H, S) are never together. That’s more complex — you’d first arrange the consonants, then place vowels in the gaps.

But for a first meeting, the pattern is always:

The gap method for "never together"

If you want A and B never adjacent, another direct way:

1. Arrange the other $n-2$ items first: $(n-2)!$ ways …