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'.
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 $=20+18+360+360=758$
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