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Question: Number of \(4\) letter words which can be formed using the alphabets of “INFINITE” is equal to: A....

Number of 44 letter words which can be formed using the alphabets of “INFINITE” is equal to:
A. 144144
B. 286286
C. 288288
D. 148148

Explanation

Solution

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. nCr=n!r!(n−r)!{}^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 44 letter words using the letters in INFINITE.Here, some of the letters are repeated in the word so we can’t simply use 8P4{}^8{P_4}. So, let's first count the letters in the word.
Number of I’s in the word ‘INFINITE’ =3 = 3
Number of N’s in the word ‘INFINITE’ =2 = 2
Number of F’s in the word ‘INFINITE’ =1 = 1
Number of T’s in the word ‘INFINITE’ =1 = 1
Number of E’s in the word ‘INFINITE’ =1 = 1
Since the letters of the word are being repeated. So, we will form cases.

Case 11: When all the 44 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 =5C4×4! = {}^5{C_4} \times 4!
Now, using the formula, nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}, we get,
5!1!×4!×4!\dfrac{{5!}}{{1! \times 4!}} \times 4!
Cancelling the common factors in numerator and denominator, we get,
Number of words=5×4×3×2×11\text{Number of words}=\dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{1}
⇒Number of words=120\Rightarrow \text{Number of words}= 120

Case 22: 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 2C1{}^2{C_1} ways and the different letters in 4C2{}^4{C_2} ways. Now, we know that if rr things are alike out of total n things can be arranged in n!r!\dfrac{{n!}}{{r!}} ways. So, they can be arranged among themselves in 4!2!\dfrac{{4!}}{{2!}} ways.So, we have,
Number of words=2C1×4C2×4!2!\text{Number of words} = {}^2{C_1} \times {}^4{C_2} \times \dfrac{{4!}}{{2!}}
⇒Number of words=2!1×4×3×2!2!×2!×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,
Number of words=1×4×31×4×3×2!2!\text{Number of words} = 1 \times \dfrac{{4 \times 3}}{1} \times \dfrac{{4 \times 3 \times 2!}}{{2!}}
⇒Number of words=4×3×4×3\Rightarrow \text{Number of words} = 4 \times 3 \times 4 \times 3
⇒Number of words=144\Rightarrow \text{Number of words} = 144

Case 33: When 22 letters are alike and the remaining 22 letters are also alike.We can take the two alike letters from the repeated letters in 2C2{}^2{C_2} ways and then they are arranged among themselves in 4!2!×2!\dfrac{{4!}}{{2! \times 2!}} ways.So, we have,
Number of words=2C2×4!2!×2!\text{Number of words} = {}^2{C_2} \times \dfrac{{4!}}{{2! \times 2!}}
⇒Number of words=2!0!×2!×4×3×2!2!×2!\Rightarrow \text{Number of words} = \dfrac{{2!}}{{0! \times 2!}} \times \dfrac{{4 \times 3 \times 2!}}{{2! \times 2!}}
⇒Number of words=1×4×32\Rightarrow \text{Number of words} = 1 \times \dfrac{{4 \times 3}}{2}
⇒Number of words=6\Rightarrow \text{Number of words} = 6

Case 44: When 33 letters are alike and 11 letters are different. We can take the three alike letters in 1C1{}^1{C_1} ways and the one different letter in 4C1{}^4{C_1} ways and then they are arranged among themselves in 4!3!\dfrac{{4!}}{{3!}}ways.So, we have,
Number of words=1C1×4C1×4!3!\text{Number of words}= {}^1{C_1} \times {}^4{C_1} \times \dfrac{{4!}}{{3!}}
⇒Number of words=1×4!3!×1!×4×3!3!\Rightarrow \text{Number of words}= 1 \times \dfrac{{4!}}{{3! \times 1!}} \times \dfrac{{4 \times 3!}}{{3!}}
⇒Number of words=4×3!3!×4\Rightarrow \text{Number of words}= \dfrac{{4 \times 3!}}{{3!}} \times 4
∴Number of words=16 \therefore \text{Number of words}=16 \\\
Total number of four digit letters =120+144+6+16=286 = 120 + 144 + 6 + 16 = 286. Hence, the number of 44 letter words which can be formed using the alphabets of word “INFINITE” is 286286.

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 nPr{}^n{P_r}, and this expression is equal to, nPr=n!(n−r)!{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}. If we try to solve the problem with8P4{}^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.