Question

How many 4 letter words can be formed using the letters of the word PROPORTION.

Answer 1

Hint: To find this question, first we need to know the exact No’s of letters given in the word PROPORTION and which may arise in a few cases. Such as: 4 distinct letters, 2 distinct letters repeated twice, exactly a letter repeating twice and exactly a letter repeating thrice, by arranging the letter in permutation and combination accordingly.

Complete step by step solution:
Here, the word given is PROPORTION in which we have:
$\begin{aligned} & P\to 2\text{ Times} \\\ & R\to 2\text{ Times} \\\ & O\to 3\text{ Times} \\\ & \text{T}\to 1\text{ Time} \\\ & \text{I}\to \text{1 Time} \\\ & \text{N}\to \text{1 Time} \\\ \end{aligned}$
To find the No.’s of 4 letter words using the letters P, R, O, T & I following cases arise:
Case 1: Word with 4 distinct letters
We have 6 letters in total to form a word with 4 letters.
So, we can arrange this letters in ${}^{6}{{P}_{4}}$ .We know that –
${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$
Here, we get –
${}^{6}{{P}_{4}}=\dfrac{6!}{\left( 6-4 \right)!}$
$\Rightarrow \dfrac{6\times 5\times 4\times 3\times 2\times 1}{2\times 1}=\dfrac{3840}{2}$
$=360\text{ ways}$ .
Case 2: Word with 2 distinct letters repeating twice.
The two letters out of three repeating letters can be selected in the form of ${}^{3}{{C}_{2}}$ . We know that –
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Here, we get –
$\begin{aligned} & {}^{3}{{C}_{2}}=\dfrac{3!}{2!\left( 3-2 \right)!} \\\ & =\dfrac{3\times 2\times 1}{2\times 1\left( 1 \right)} \\\ \end{aligned}$
$=3\text{ ways}$ .
Now, each combination can be arranged in –
$=\dfrac{4!}{2!\times 2!}$
$=\dfrac{4\times 3\times 2\times 1}{2\times 2}$
$=6\text{ ways}$ .
So, total No.’s of such words $=3\times 6=18$
Case 3: Words with exactly a letter repeating twice.
The repeating letters are P, R & O. So, we will choose one of these letters in the form of ${}^{3}{{C}_{1}}$ .
Here,
$\begin{aligned} & {}^{3}{{C}_{1}}=\dfrac{3!}{1!\left( 3-1 \right)!} \\\ & =\dfrac{3\times 2\times 1}{2\times 1} \\\ \end{aligned}$
$=3\text{ ways}$
The other two distinct letter can be selected in ${}^{5}{{C}_{2}}$ . we get –
$\begin{aligned} & {}^{5}{{C}_{2}}=\dfrac{5!}{2!\left( 5-2 \right)!} \\\ & =\dfrac{5\times 4\times 3\times 2\times 1}{2\times 3\times 2\times 1} \\\ \end{aligned}$
$=10\text{ ways}$
Now, each combination can be arranged in –
$=\dfrac{4!}{2!}$
$\begin{aligned} & =\dfrac{4\times 3\times 2\times 1}{2} \\\ & =12\text{ ways} \\\ \end{aligned}$
So, total No. of such words is –
$\begin{aligned} & =3\times 10\times 12 \\\ & =360 \\\ \end{aligned}$
Case 4: Words with exactly a letter repeating thrice.
We have only one letter which repeats thrice i.e. ‘O’.
Now, we have to select 1 letter out of the remaining options.
So, we can arrange it as:${}^{5}{{P}_{1}}$
Here, ${}^{5}{{P}_{1}}=5\text{ ways}$ .
Now, each combination can be arranged in $\dfrac{4!}{3!}$ .
Here, we get –
$\dfrac{4!}{3!}=\dfrac{4\times 3\times 2\times 1}{3\times 2\times 1}$
$=4\text{ ways}$
So, total No.’s of such words is –
$\Rightarrow 5\times 4=20$
Therefore, all possible No.’s of arrangements is –
$\Rightarrow 360+360+18+20=758\text{ ways}$
Hence, 758 ways of 4 letter words can be arranged by the letters of the word PROPORTION.

Note: Generally students get confused between combination & permutation. If you have to select use combination ${}^{n}{{C}_{r}}$ and if you have to arrange use permutation ${}^{n}{{P}_{r}}$ . it is very nice trick to use.
Don’t forget to consider all possibilities or else you might get the wrong answer. For example: if you missed any of the situation/case then you will get the wrong answer.