Question

Find the number of ways in which
(ii) an arrangement of four letters can be made from the letters of the word 'PROPORTION'

Answer 1

Finding the number of arrangements by making cases
As there are 10 letters in the word PROPORTION,
i.e., OOO, PP, RR, I, T and N.
We have to form four-letter words.
We will discuss all possible cases one by one.

Case (I): 3 alike letters and 1 distinct letter:
There are 3 alike letters, OOO, which can be selected by 1 way. And Out of the 5 different letters, P, R, I, T and N, one can be selected by
${}^5{C}_1=5\text{ways.}$
Letters can be arranged by $\frac{4!}{3!1!}$ ways.
So, total number of ways
$={}^5{C}_1\times \frac{4!}{3!1!}=5\times 4=20$

Case (II): 2 alike of one kind and 2 alike of other kind.
There are 3 groups of two alike letters, so it can be selected by ${}^3{C}_2=3\text{ways}$.
4 letters can be arranged by $\frac{4!}{2!2!}$ ways.
So, total number of ways
$={}^3{C}_2\times \frac{4!}{2!2!}=3\times 6=18$

Case (III): 2 alike letters and 2 different letters.
There are 3 pair of two alike letters of which one pair can be selected in ${}^3{C}_1$ ways.
Now, from the remaining 5 letters, 2 different letters can be chosen by ${}^5{C}_2$ ways.
So, 2 alike letters and 2 different letters can be selected by ${}^3{C}_1\times {}^5{C}_2=3\times 10=30\text{ways}$
Letters can be arranged by $\frac{4!}{2!}\text{ways}$
So, total number of ways
$=30\times \frac{4!}{2!}=30\times 12=360$

Case (IV): 4 different letters.
There are 6 different letters. So, number of ways of selecting 4 letters
$={}^6{C}_4=\frac{6!}{4!2!}=\frac{6\times 5\times 4!}{4!2!}=15$
4 letters can be arranged by 4! ways.
So, total number of ways
$=15\times 4!=15\times 24=360$

Hence, the total number of arrangements
$=20+18+360+360$
$=758$