Question

Number of $4$ letter words which can be formed using the alphabets of “INFINITE” is equal to:
A. $144$
B. $286$
C. $288$
D. $148$

Answer 1

Hint: The given question involves the concepts of permutations and combinations. We are required to find the number of four letter words using the alphabet of the word “INFINITE”. First we count the number of different letters in the word “INFINITE”, then we will make cases if the letters are repeated or not then by using the formula of combination i.e. ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$. The cases will be that all the four letters are distinct, the second case is when two letters are same and the other two are distinct, the third case is when all the four letters are from the repeated letters as two from one letter and two from other, and the fourth case is that all letters are repeated as three letters are from one alphabet and one from other. We find the number of possibilities in that particular case. At last, we’ll add the result of all the cases to get the answer.

Complete step by step answer:
We are given the word INFINITE. We are to find $4$ letter words using the letters in INFINITE.Here, some of the letters are repeated in the word so we can’t simply use ${}^8{P_4}$. So, let's first count the letters in the word.
Number of I’s in the word ‘INFINITE’ $ = 3$
Number of N’s in the word ‘INFINITE’ $ = 2$
Number of F’s in the word ‘INFINITE’ $ = 1$
Number of T’s in the word ‘INFINITE’ $ = 1$
Number of E’s in the word ‘INFINITE’ $ = 1$
Since the letters of the word are being repeated. So, we will form cases.

Case $1$: When all the $4$ letters are distinct. So, we have to choose four letters out of the five different letters from the word ‘INFINITE’ and then arrange them. So, we get,
The number of words $ = {}^5{C_4} \times 4!$
Now, using the formula, ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, we get,
$\dfrac{{5!}}{{1! \times 4!}} \times 4!$
Cancelling the common factors in numerator and denominator, we get,
$\text{Number of words}=\dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{1}$
$\Rightarrow \text{Number of words}= 120$

Case $2$: When two letters are repeated. We have to make four letter words. So, two letters in the word are alike and the rest two are different from the non-repeated letters. So, we can choose the letters in ${}^2{C_1}$ ways and the different letters in ${}^4{C_2}$ ways. Now, we know that if $r$ things are alike out of total n things can be arranged in $\dfrac{{n!}}{{r!}}$ ways. So, they can be arranged among themselves in $\dfrac{{4!}}{{2!}}$ ways.So, we have,
$\text{Number of words} = {}^2{C_1} \times {}^4{C_2} \times \dfrac{{4!}}{{2!}}$
$\Rightarrow \text{Number of words} = \dfrac{{2!}}{1} \times \dfrac{{4 \times 3 \times 2!}}{{2! \times 2!}} \times \dfrac{{4!}}{{2!}}$
Simplifying the calculations, we get,
$\text{Number of words} = 1 \times \dfrac{{4 \times 3}}{1} \times \dfrac{{4 \times 3 \times 2!}}{{2!}}$
$\Rightarrow \text{Number of words} = 4 \times 3 \times 4 \times 3$
$\Rightarrow \text{Number of words} = 144$

Case $3$: When $2$ letters are alike and the remaining $2$ letters are also alike.We can take the two alike letters from the repeated letters in ${}^2{C_2}$ ways and then they are arranged among themselves in $\dfrac{{4!}}{{2! \times 2!}}$ ways.So, we have,
$\text{Number of words} = {}^2{C_2} \times \dfrac{{4!}}{{2! \times 2!}}$
$\Rightarrow \text{Number of words} = \dfrac{{2!}}{{0! \times 2!}} \times \dfrac{{4 \times 3 \times 2!}}{{2! \times 2!}}$
$\Rightarrow \text{Number of words} = 1 \times \dfrac{{4 \times 3}}{2}$
$\Rightarrow \text{Number of words} = 6$

Case $4$: When $3$ letters are alike and $1$ letters are different. We can take the three alike letters in ${}^1{C_1}$ ways and the one different letter in ${}^4{C_1}$ ways and then they are arranged among themselves in $\dfrac{{4!}}{{3!}}$ways.So, we have,
$\text{Number of words}= {}^1{C_1} \times {}^4{C_1} \times \dfrac{{4!}}{{3!}}$
$\Rightarrow \text{Number of words}= 1 \times \dfrac{{4!}}{{3! \times 1!}} \times \dfrac{{4 \times 3!}}{{3!}}$
$\Rightarrow \text{Number of words}= \dfrac{{4 \times 3!}}{{3!}} \times 4$
$\therefore \text{Number of words}=16 \\ $
Total number of four digit letters $ = 120 + 144 + 6 + 16 = 286$. Hence, the number of $4$ letter words which can be formed using the alphabets of word “INFINITE” is $286$.

So, option B is the correct answer.

Note: A permutation is selecting all the ordered pair of ‘r’ elements out of ‘n’ total elements is given by ${}^n{P_r}$, and this expression is equal to, ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$. If we try to solve the problem with${}^8{P_4}$, then we end up considering the similar letters as distinct letters and will get more than the possible cases, as for example “N” and “N” are same, so if a word is formed “NFNE” and “NFNE” they will be same, as both the N’s are same. Therefore, we must take care of the language and calculations in such questions.