Q. 6

# Find the nu

Given that we need to find the no. of permutations formed by r things which were taken from n distinct things in which 2 specified things must occur together.

Here, it is clear that 2 things are already selected and we need to choose (r – 2) things from the remaining (n – 2) things.

Let us find the no. of ways of choosing (r – 2) things and assume it to be N1.

N1 = (No. of ways of choosing (r – 2) things from remaining (n – 2) things)

N1 = n – 2Cr – 2

Now we need to find the no. of permutations than can be formed using 2 things which are together.

Now we need to arrange the chosen 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 2!.

Now let us assume the together things as a single thing this gives us total (r – 1) things which were present now.

Now, we need to arrange these (r – 1) 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 – 1)!.

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

N = N1 × 2! × (r – 1)!

N = n – 2Cr – 2 × 2 × (r – 1)!

The no. of permutations that can be formed by r things which are chosen from n things in which 3 things are always together are n – 2Cr – 2 × 2 × (r – 1)!.

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