**Problem: **Show that the number of ways in which four distinct integers can be chosen from 1,2,…n \((n \ge 7) \) such that no two are consecutive is equal to \(\displaystyle{\binom{n-3}{4}} \).

**Solution:** suppose we chose a>b>c>d four integers from 1,2,…n such that no two are consecutive.So d<c-1<b-2<a-3.

Now d,(c-1),(b-2),(a-3) are four distinct integers from 1,2,…(n-3),which may be consecutive or not.

So number of ways to chose four numbers such that no two are consecutive from 1,2,…n = number of ways to chose four distinct numbers from 1,2,…(n-3).

Now number of ways to chose 4 ways from 1,2,…(n-3) = \(\displaystyle{\binom{n-3}{4}} \).

**Conclusion: ** Number of ways to chose four distinct integers from 1,2,…n such that no two are consecutive is \(\displaystyle{\binom{n-3}{4}} \)

*Related*

## No comments, be the first one to comment !