Question

Find the number of ways in which selection of four words from proportion

Answer 1

First of all see that which letters are repeating.

We have two P’s,two R’s,three O’s and all the other,i.e,T,I and N have appeared for once.

Now,the cases are:-

1.Words with four distinct letters.

We have 6 letters all total,i.e,(I,N,P,R,O and T)so we can arrange this letters in (64)×4!=360ways.

2.Words with exactly a letter repeated twice.

We have P,R and O repeating itself.Now one of this three letter can be choose in (31)=3ways.

The other two distinct letters can be selected in (52)=10ways

Now each combination can be arranged in 4!2!=12ways

So,total no. of such words=3×10×12=360

3.Words with exactly two distinct letters repeated twice.

Two letters out of the three repeating letters P,R and O can be selected in (32)=3ways

Now each combination can be arranged in 4!2!×2!=6

So,total no. of such words=3×6=18

4.Words with exactly a letter repeated thrice.

We have only one option for this as our main letter that is O.

Now we have to select 1 letter out of the 5 remaining options so no. of ways to this =(51)=5

Now each combination can be arranged in 4!3!=4

So,total no. of such words=1×5×4=20

So,all possible no. of arrangements =360+360+18+20=758ways