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

Given the word is

PROPORTION

When

Number of P is $=2$

Number of O is $=3$

Number of R is $=2$

Number of I is $=1$

Number of T is $=1$

Number of N is $=1$

Let us frist find the number of selections.

Case 1:- All four letters distinct = number of ways will be $4{\times }^6{C}_4=4!\times \frac{6!}{4!(6-4)!}=\frac{6\times 5\times 4\times 3\times 2\times 1}{2!}=360$

Case 2:- Three letters same and one letter distinct, then no. of ways = ${\frac{4!}{3!}}^5{C}_1=\frac{4\times 3!}{3!}\times 5=20$

Case 3:- Two letters of one type and the other two letters of other type, then no. of ways(out for P,R and O)=${}^3{C}_2{\times }^4{C}_2=3\times \frac{4!}{2!\times 2!}=18$

Case 4:-two letters same and other two letters are different.then, no. of ways. = ${}^3{C}_1{\times }^5{C}_2=3\times 10=30\times \frac{4!}{2!}=360$

Now according to given question,

Part (1):- Total number of selections=$15+5+3+30=53$ Ans.

Part (1):- Total arrangement $=360+20+18+360=758$