Question

Find the number of permutations in the letters of the word PROPORTION taken $4$ at a time.

Answer 1

We have two P's, two R's, three O's and other letters T, I, and N have appeared for once

1. Words with four distinct letters.

We have 6 letters all total (I,N,P,R,O and T) so we can arrange this letter in ${}^6{C}_4\times 4!=\underset{–}{360}$ ways

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 ${}^3{C}_1=3$ ways

The other two distinct letters can be selected in ${}^5{C}_2=10$ ways.

Now each combination can be arranged in $\frac{4!}{2!}=12$ ways.

So total number of such words $3\times 10\times 12=\underset{–}{360}$

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 ${}^3{C}_2=3$ ways.

Now each combination can be arranged in $\frac{4!}{2!\times 2!}=6$

So, total number of such words $=3\times 6=\underset{–}{18}$

4. Words with exactly a letter repeated thrice

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

Now we have to select 1 letter out of the 5 remaining options so number of ways to select this ${}^5{C}_1=5$ ways

Now each combination can be arranged in $\frac{4!}{3!}=4$

So, total number of such words $=5\times 4=\underset{–}{20}$

So, all possible number of arrangements $=360+360+18+20=758ways$