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Question: Find the number of three letters words that can be formed from the word ‘SERIES’....

Find the number of three letters words that can be formed from the word ‘SERIES’.

Explanation

Solution

Hint: In this particular type of question use the concept that if there are n different objects and we have to select r different objects. So the number of ways to select and arrange them is nPr{}^n{P_r}. So use this concept to reach the solution of the question.

Complete step-by-step answer:
Given word-
SERIES
So as we see that there are a total 6 letters in the given word but only 4 letters are different.
Now as we know if there are n different objects and we have to select r different objects so the number of ways to select and arrange them = nPr{}^n{P_r}
Case – 1
So the number of ways to select and arrange 3 letters from the given 4 letters (all are different) = 4P3{}^4{P_3}.
Case – 2
Now consider the case when we choose two same letters and one different letter, as there are two letters S and E which are repeated so the number of ways to select any one of the repeated letter = 2C1{}^2{C_1}, as the letter is same so there are no arrangements.
Now we have to write one more letter so that the 3 letter word is completed.
So the remaining letters in the given word = (6 – 2) = 4.
In these remaining letters different letters are (4 – 1) = 3.
So the number of ways to select one letter out of 3 is 3C1{}^3{C_1}
Now we choose two same letters and one different letter. Now we have to arrange them.
So the number of ways to arrange them = 3!2!\dfrac{{3!}}{{2!}} (divide by 2! Is because there are two same letters).
Now the total number of ways to select one pair of letters which is the same and one different letter from the given letters is the multiplication of the above values.
⇒2C1×3C1×3!2!\Rightarrow {}^2{C_1} \times {}^3{C_1} \times \dfrac{{3!}}{{2!}}
So the number of three letters words that can be formed from the word ‘SERIES’ = 4P3{}^4{P_3} + 2C1×3C1×3!2!{}^2{C_1} \times {}^3{C_1} \times \dfrac{{3!}}{{2!}}
Now simplify this according to property, nPr=n!(n−r)!,nCr=n!r!(n−r)!{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}},{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} we have,
Total number of three letter words = 4!(4−3)!+2!1!(2−1)!×3!1!(3−1)!×3.2!2!\dfrac{{4!}}{{\left( {4 - 3} \right)!}} + \dfrac{{2!}}{{1!\left( {2 - 1} \right)!}} \times \dfrac{{3!}}{{1!\left( {3 - 1} \right)!}} \times \dfrac{{3.2!}}{{2!}}
Total number of three letter words = 4.3.2+2×3×3=24+18=424.3.2 + 2 \times 3 \times 3 = 24 + 18 = 42
So three are a total of 42 three letter words.
So this is the required answer.

Note – Whenever we face such types of questions the key concept we have to remember is that if there are n objects and in n objects p objects are of same kind and q objects are of another same kind then the number of ways to arrange them is given as, n!p!q!\dfrac{{n!}}{{p!q!}}, where, n > (p + q) .