Question

The number of permutations of n things taken “r” at a time if 3 particular things always occur is:
$
  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)}}! \\
 $

Answer 1

Hint: 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 –
$P(n,r) = \dfrac{{n!}}{{\left( {n - r} \right)!}}$
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 – $^{n - s}{C_{r - s}} \times r!$.
In this question, we need to group the three things together so that the total number of things reduced to $(n - 3)$ 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 $26 - 3 = 23$ letters, and we have to choose $10 - 3 = 7$ letters out of them as we have already chosen 3 letters. The r things can also arrange themselves in $r!$ ways.
Similarly, we have to choose $\left( {r - 3} \right)$ things from$\left( {n - 3} \right)$things that can be done in $^{\left( {n - 3} \right)}{C_{r - 3}}$ways.
Now, r things can also arrange themselves in $r!$ 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)}{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 $\dfrac{{\left( {n - 3} \right)!}}{{\left( {n - r} \right)!}} \cdot r(r - 1)(r - 2)$.

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$^{\left( {n - 3} \right)}{C_{r - 3}} \times r!$. But note that$^{\left( {n - 3} \right)}{C_{r - 3}} \times r! = r{ \times ^{\left( {n - 3} \right)}}{P_{r - 3}}$. You can easily find the relation if you know the permutation and combination formula.