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Question: The number of permutations of n things taken “r” at a time if 3 particular things always occur is: ...

The number of permutations of n things taken “r” at a time if 3 particular things always occur is:
A)(n−3)!(n−r)!⋅r(r−1)(r−2) B)(n−3)!(r−3)! C)3⋅(n−3)(n−r) D)(n−3)(r−2)!  A) \dfrac{{\left( {n - 3} \right)!}}{{\left( {n - r} \right)!}} \cdot r(r - 1)(r - 2) \\\ B) \dfrac{{(n - 3)!}}{{(r - 3)!}} \\\ C) 3 \cdot \dfrac{{(n - 3)}}{{(n - r)}} \\\ D) \dfrac{{(n - 3)}}{{(r - 2)}}! \\\

Explanation

Solution

Permutation of a set is an arrangement of its members into a sequence or linear order. The number of permutation of ‘n’ object taken ‘r’ at a time is given by –
P(n,r)=n!(n−r)!P(n,r) = \dfrac{{n!}}{{\left( {n - r} \right)!}}
This is the general formula. But the number of permutations of n distinct things taken r at a time when particular things are always included in each arrangement is given as – n−sCr−s×r!^{n - s}{C_{r - s}} \times r!.
In this question, we need to group the three things together so that the total number of things reduced to (n−3)(n - 3) and those 3 things can be arranged in 3! ways.

Complete step-by-step solution:
Here we have ‘n’ things, and we need to find out the permutation of ‘r’ things taken at a time such that 3 things always occur.
If we take an example, suppose we have letters from A to Z. Here, we have to choose10 letters such that A, B, and C always occur, so now we have total 26−3=2326 - 3 = 23 letters, and we have to choose 10−3=710 - 3 = 7 letters out of them as we have already chosen 3 letters. The r things can also arrange themselves in r!r! ways.
Similarly, we have to choose (r−3)\left( {r - 3} \right) things from(n−3)\left( {n - 3} \right)things that can be done in (n−3)Cr−3^{\left( {n - 3} \right)}{C_{r - 3}}ways.
Now, r things can also arrange themselves in r!r! ways.
The number of permutations of n things taken ‘r’ at a time if 3 particular things always occur is given as:
P=(n−3)Cr−3×r! =n−3![n−3−(r−3)]!(r−3)!r! =(n−3)!r(r−1)(r−2)(r−3)!(n−r)!(r−3)! =(n−3)!r(r−1)(r−2)(n−r)!  P = {}^{(n - 3)}{C_{r - 3}} \times r! \\\ = \dfrac{{n - 3!}}{{[n - 3 - \left( {r - 3} \right)]!(r - 3)!}}r! \\\ = \dfrac{{(n - 3)!r(r - 1)(r - 2)(r - 3)!}}{{\left( {n - r} \right)!(r - 3)!}} \\\ = \dfrac{{(n - 3)!r(r - 1)(r - 2)}}{{\left( {n - r} \right)!}} \\\
Hence, the number of permutations of n things taken ‘r’ at a time if 3 particular things always occur is (n−3)!(n−r)!⋅r(r−1)(r−2)\dfrac{{\left( {n - 3} \right)!}}{{\left( {n - r} \right)!}} \cdot r(r - 1)(r - 2).

Hence the correct answer is option (A).

Note: Students might find a problem in understanding the formula where we found the number of permutation in terms of(n−3)Cr−3×r!^{\left( {n - 3} \right)}{C_{r - 3}} \times r!. But note that(n−3)Cr−3×r!=r×(n−3)Pr−3^{\left( {n - 3} \right)}{C_{r - 3}} \times r! = r{ \times ^{\left( {n - 3} \right)}}{P_{r - 3}}. You can easily find the relation if you know the permutation and combination formula.