The total number of ways of selecting five letters from the letters of the word INDEPENDENT is
(A) $ 72 $
(B) $ 12 $
(C) $ 24 $
(D) $ 48 $
Answer
623.1k+ views
Hint:
For answering this question we will first count the number of letters present in the word INDEPENDENT and their repetitions after that we will find the number of ways in which the 5 letters can be selected and calculate the number of ways of each arrangement possible and then sum them all.
Complete step by step answer:
Now considering the question the word given to us is INDEPENDENT which has I,P,T,2 D’s , 3 E’s, and 3 N’s that is 6 letters in which some are repeated.
We need to find the total number of ways of selecting five letters from the letters of the given word INDEPENDENT which has I,P,T,2 D’s , 3 E’s and 3 N’s that is 6 letters in which some are repeated.
The arrangements of these 5 letters can be done in any way we have no conditions here. So while arranging the selected letters we should consider all the possible ways. As here we have 6 letters in which one letter is repeated twice and two letters are repeated thrice and three letters occur only one time and we need to select 5 letters. So, we can select 5 different letters among the 6 letters and we can select 5 letters in which some letters are alike for maximum there will be 3 same letters because at most one letter has occurred thrice only in the word.
Hence the 5 letters can be selected in the following ways:
(i) When out of 5 letters 3 letters alike and the remaining 2 different letters, the number of ways will be
$ \begin{align}
& ^{2}{{C}_{1}}{{\times }^{5}}{{C}_{2}}=2\times \left( \dfrac{5!}{2!\times 3!} \right) \\
& \Rightarrow 2\times \left( \dfrac{5\times 4}{2} \right)=20 \\
\end{align} $ .
As here we have 2 letters which are repeated thrice so we have to select one letter among them so we will use $ ^{2}{{C}_{1}} $ and for the remaining 2 different letters we can select any 2 from the 5 remaining letters so we will use $ ^{5}{{C}_{2}} $ .
(ii) When out of 5 letters 3 letters alike and other 2 letters alike, the number of ways will be $ \begin{align}
& ^{2}{{C}_{1}}{{\times }^{2}}{{C}_{1}}=2\times 2 \\
& \Rightarrow 4 \\
\end{align} $ .
As here we have 2 letters which are repeated thrice so we have to select one letter among them so we will use $ ^{2}{{C}_{1}} $ and for the remaining 2 letters we can select any 1 from the 2 repeating letters so we will use $ ^{2}{{C}_{1}} $ .
(iii) When out of 5 letters 2 letters alike and the remaining 3 different letters, the number of ways will be
$ \begin{align}
& ^{3}{{C}_{1}}{{\times }^{5}}{{C}_{3}}=3\times \left( \dfrac{5!}{3!\times 2!} \right) \\
& \Rightarrow 3\times \left( \dfrac{5\times 4}{2} \right)=30 \\
\end{align} $ .
As here we have 3 letters which are getting repeated so we have to select one letter among them so we will use $ ^{3}{{C}_{1}} $ and for the remaining 3 different letters we can select any 3 from the 5 remaining letters so we will use $ ^{5}{{C}_{3}} $ .
(iv)When out of 5 letters 2 letters alike and other 2 letters alike and the remaining 1 different letter, the number of ways will be
$ \begin{align}
& ^{3}{{C}_{2}}{{\times }^{4}}{{C}_{1}}=3\times 4 \\
& \Rightarrow 12 \\
\end{align} $ .
As here we have 3 letters which are getting repeated so we have to select 2 letters among them so we will use $ ^{3}{{C}_{2}} $ and for the remaining one letter we can select any 1 from the 4 remaining letters so we will use $ ^{4}{{C}_{1}} $ .
(v) When out of 5 letters all letters are different, the number of ways will be $ ^{6}{{C}_{5}}=6 $ .
As here we have 6 letters in total and we have to select 5 letters among them so we will use $ ^{6}{{C}_{5}} $ .
Therefore the total number of ways of selecting five letters from the letters of the word INDEPENDENT is the sum of all of them which is given as $ 20+4+30+12+6=72 $ .
Hence option A is correct.
Note:
While answering questions of this type we should be careful while arranging the selection of 5 letters in the possible ways and calculating the number of ways. If we had forgotten one arrangement possible during the selection like we had forgotten to consider the selection of 5 different letters then we will end up having the total number of ways of selections as $ 66 $ which we can’t find in the options which implies that it is a wrong answer.
For answering this question we will first count the number of letters present in the word INDEPENDENT and their repetitions after that we will find the number of ways in which the 5 letters can be selected and calculate the number of ways of each arrangement possible and then sum them all.
Complete step by step answer:
Now considering the question the word given to us is INDEPENDENT which has I,P,T,2 D’s , 3 E’s, and 3 N’s that is 6 letters in which some are repeated.
We need to find the total number of ways of selecting five letters from the letters of the given word INDEPENDENT which has I,P,T,2 D’s , 3 E’s and 3 N’s that is 6 letters in which some are repeated.
The arrangements of these 5 letters can be done in any way we have no conditions here. So while arranging the selected letters we should consider all the possible ways. As here we have 6 letters in which one letter is repeated twice and two letters are repeated thrice and three letters occur only one time and we need to select 5 letters. So, we can select 5 different letters among the 6 letters and we can select 5 letters in which some letters are alike for maximum there will be 3 same letters because at most one letter has occurred thrice only in the word.
Hence the 5 letters can be selected in the following ways:
(i) When out of 5 letters 3 letters alike and the remaining 2 different letters, the number of ways will be
$ \begin{align}
& ^{2}{{C}_{1}}{{\times }^{5}}{{C}_{2}}=2\times \left( \dfrac{5!}{2!\times 3!} \right) \\
& \Rightarrow 2\times \left( \dfrac{5\times 4}{2} \right)=20 \\
\end{align} $ .
As here we have 2 letters which are repeated thrice so we have to select one letter among them so we will use $ ^{2}{{C}_{1}} $ and for the remaining 2 different letters we can select any 2 from the 5 remaining letters so we will use $ ^{5}{{C}_{2}} $ .
(ii) When out of 5 letters 3 letters alike and other 2 letters alike, the number of ways will be $ \begin{align}
& ^{2}{{C}_{1}}{{\times }^{2}}{{C}_{1}}=2\times 2 \\
& \Rightarrow 4 \\
\end{align} $ .
As here we have 2 letters which are repeated thrice so we have to select one letter among them so we will use $ ^{2}{{C}_{1}} $ and for the remaining 2 letters we can select any 1 from the 2 repeating letters so we will use $ ^{2}{{C}_{1}} $ .
(iii) When out of 5 letters 2 letters alike and the remaining 3 different letters, the number of ways will be
$ \begin{align}
& ^{3}{{C}_{1}}{{\times }^{5}}{{C}_{3}}=3\times \left( \dfrac{5!}{3!\times 2!} \right) \\
& \Rightarrow 3\times \left( \dfrac{5\times 4}{2} \right)=30 \\
\end{align} $ .
As here we have 3 letters which are getting repeated so we have to select one letter among them so we will use $ ^{3}{{C}_{1}} $ and for the remaining 3 different letters we can select any 3 from the 5 remaining letters so we will use $ ^{5}{{C}_{3}} $ .
(iv)When out of 5 letters 2 letters alike and other 2 letters alike and the remaining 1 different letter, the number of ways will be
$ \begin{align}
& ^{3}{{C}_{2}}{{\times }^{4}}{{C}_{1}}=3\times 4 \\
& \Rightarrow 12 \\
\end{align} $ .
As here we have 3 letters which are getting repeated so we have to select 2 letters among them so we will use $ ^{3}{{C}_{2}} $ and for the remaining one letter we can select any 1 from the 4 remaining letters so we will use $ ^{4}{{C}_{1}} $ .
(v) When out of 5 letters all letters are different, the number of ways will be $ ^{6}{{C}_{5}}=6 $ .
As here we have 6 letters in total and we have to select 5 letters among them so we will use $ ^{6}{{C}_{5}} $ .
Therefore the total number of ways of selecting five letters from the letters of the word INDEPENDENT is the sum of all of them which is given as $ 20+4+30+12+6=72 $ .
Hence option A is correct.
Note:
While answering questions of this type we should be careful while arranging the selection of 5 letters in the possible ways and calculating the number of ways. If we had forgotten one arrangement possible during the selection like we had forgotten to consider the selection of 5 different letters then we will end up having the total number of ways of selections as $ 66 $ which we can’t find in the options which implies that it is a wrong answer.
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