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'.

Answer 1

The given word is P R O P O R T I O N.

Total letters = 10

Number of P = 2, Number of R = 2

Number of O = 3, Number of T = 1

Number of P = 1, Number of N = 1

(i) Case I : There are 6 different letters is which all the four are distinct to selected.

Number of ways to select therefore ${=}^6{C}_4=15$

(i) Case II : Two same and two distinct letters are selected there are three pairs which more than, letters.

Number of ways to select therefore

${=}^3{C}_1{\times }^5{C}_2$

$=3\times 10$

$=30$

Case III

Two alike of one kind and two alike of other kind.

There are 3 pairs of letters is more than one letters. Any 2 of these 3 letters.

Number of ways to select these letters

${=}^3{C}_2$

$=3$

Case IV : Three alike and one different. Number of ways to select these letters

$=1{\times }^5{C}_1$

$=5$

Therefore,

Number of ways to select four letters

= 15 + 30 + 3 + 5

= 53

Required number of ways to select = 53

(ii) For case I :

Number of arrangements of four letters all distrinct

${=}^6{C}_4\times 4!$

$=15\times 24$

$=360$

For case II :

Number of arrangements of four letters two same kind and two of different kind

$={}^3{C}_1\times {}^5{C}_2\times \frac{4!}{2!1!1!}$

$=3\times 10\times 12$

$360$

For Case III :

Number of arrangements of four letters two alike of one kind and two of other kind

$={}^3{C}_2\times \frac{4!}{2!2!}$

$=3\times 6$

$=18$

Case IV :

Number of arrangements of four letters 3 alike and 1 other kind $1{\times }^5{C}_1\times \frac{4!}{3!1!}=20$

Therefore,

Total number of arrangements of four letters selected = 360+360+18+20

Required number of arrangement = 758