Given that we need to find the no. of permutations formed by r things which were taken from n distinct things in which 3 particular things must occur together.

Here, it is clear that 3 things are already selected and we need to choose (r – 3) things from the remaining (n – 3) things.

Let us find the no. of ways of choosing (r – 3) things and assume it to be N₁.

⇒ N₁ = (No. of ways of choosing (r – 3) things from remaining (n – 3) things)

⇒ N₁ = ^{n – 3}C_{r – 3}

Now we need to find the no. of permutations than can be formed using 3 things which are together.

Now we need to arrange the chosen 3 things. Since every thing differs from other. The arrangement is similar to that of arranging n people in n places which are n! ways to arrange. So, the total no. of words that can be formed is 3!.

Now let us assume the together things as a single thing this gives us total (r – 2) things which were present now.

Now, we need to arrange these (r – 2) things. Since every thing differs from other. The arrangement is similar to that of arranging n people in n places which are n! ways to arrange. So, the total no. of words that can be formed is (r – 2)!.

Let us the total no. of words formed be N.

⇒ N = N₁ × 3! × (r – 2)!

⇒ N = ^{n – 3}C_{r – 3} × 3! × (r – 2)!

∴ The no. of permutations that can be formed by r things which are chosen from n things in which 3 things are always together is ^{n – 3}C_{r – 3} × 3! × (r – 2)!.