**Question:**

Find the number of ways in which : (a) a selection (b) an arrangement, of four letters can be made from the letters of the word 'PROPORTION'.

**Solution:**

There are 10 letters in the word PROPORTION, namely OOO, PP, RR, I, T and N.(a) The four-letter word may consists of

(i) 3 alike letters and 1 distinct letter

(ii) 2 alike letters of one kind and 2 alike letters of the second kind

(iii) 2 alike letters and 2 distinct letters

(iv) all distinct letters

Now, we shall discuss these four cases one by one.

(i) 3 alike letters and 1 distinct letter:

There is one set of three alike letters, OOO, which can be selected in one way.

Out of the 5 different letters, $P, R, I, T$ and $N$, one can be selected in $C_{1}=5$ ways.

(ii) There are 3 sets of two alike letters, which can be selected in 3C2 = 3 ways.

(iii) There are three sets of two alike letters, which can be selected in 3C1 ways.

Now, from the remaining 5 letters, 2 letters can be chosen in 5C2 ways.

Thus, 2 alike letters and 2 different letters can be selected in ${ }^{3} \mathrm{C}_{1} \times{ }^{5} \mathrm{C}_{2}=30$ ways.

(iv) There are 6 different letters.

Number of ways of selecting 4 letters = 6C4 = 15

∴ Total number of ways = 5+ 3 + 30 + 15 = 53

(b) The four-letter word may consists of

(i) 3 alike letters and 1 distinct letter

(ii) 2 alike letters of one kind and 2 alike letters of the second kind

(iii) 2 alike letters and 2 distinct letters

(iv) all distinct letters

Now, we shall discuss these four cases one by one.

(i) 3 alike letters and 1 distinct letter:

There is one set of three alike letters, OOO, which can be selected in one way.

Out of the 5 different letters, $P, R, I, T$ and $N$, one can be selected in $^{5} C_{1}$ ways.

These four letters can be arranged in $\frac{4 !}{3 ! 1 !}$ ways.

$\therefore$ Total number of ways $={ }^{5} C_{1} \times \frac{4 !}{3 ! 1 !}=20$

(ii) There are 3 sets of two alike letters, which can be selected in 3C2 ways.

Now, the letters of each group can be arranged in $\frac{4 !}{2 !} 2 !$ ways.

$\therefore$ Total number of ways $={ }^{3} C_{2} \times \frac{4 !}{2 ! 2 !}=18$

(iii) There are three sets of two alike letters, which can be selected in 3C1 ways.

Now, from the remaining 5 letters, 2 letters can be chosen in 5C2 ways.

Thus, 2 alike letters and 2 different letters can be selected in ${ }^{3} \mathrm{C}_{1} \times{ }^{5} \mathrm{C}_{2}=30$ ways.

Now, the letters of each group can be arranged in $\frac{4 !}{2 !}$ ways.

$\therefore$ Total number of ways $=30 \times \frac{4 !}{2 !}=360$

(iv) There are 6 different letters.

So, the number of ways of selecting 4 letters is 6C4 = 15 and these letters can be arranged in 4! ways.

$\therefore$ Total number of ways $=15 \times 4 !=360$

$\therefore$ Total number of ways $=20+18+360+360=758$