
The total number of three-letter words that can be formed from the letter of the word SAHARANPUR is equal to
(a) 210
(b) 237
(c) 247
(d) 227
Answer
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Hint: In this question, we need to consider all the possibilities. Now, calculate the number of 3 letter words that can be formed with all the letters being different using the permutation formula given by \[{}^{n}{{P}_{r}}\]. Then find the number of words having two letters similar using the combinations given by the formula \[{}^{n}{{C}_{r}}\] and then arrange them. Then find the words with all the 3 letters the same and add all these to get the result.
Complete step-by-step answer:
Now, from the given word SAHARANPUR in the question we have
S, A, A, A, H, R, R, N, P, U
Now, we need to find the 3 letter words that can be formed using these letters
Let us first consider the case that all the 3 letters to be different
Now, we have to arrange the 7 different letters in 3 places
As we already know that arrangement of n things in r places can be done using the permutations given by the formula
\[{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}\]
Now, on comparing with the formula we have
\[n=7,r=3\]
Now, on substituting the respective values we get,
\[\Rightarrow {}^{7}{{P}_{3}}\]
Now, this can be further written as
\[\Rightarrow \dfrac{7!}{\left( 7-3 \right)!}\]
Now, on further simplification we get,
\[\begin{align}
& \Rightarrow 7\times 6\times 5 \\
& \Rightarrow 210 \\
\end{align}\]
Thus, 210 3- letter words can be formed with all letters being different
Now, let us find the number of 3-letter words having two letters same
Here, the possible two same letters can be either A or R
As we already know that selection can be done using the combinations given by the formula
\[{}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}\]
Now, we can select either of A or R in
\[\Rightarrow {}^{2}{{C}_{1}}=\dfrac{2!}{1!1!}\]
That means in 2 ways
Now, the next letter can be any of the remaining 6 letters which can be done in
\[\Rightarrow 6\text{ ways}\]
Now, the arrangement of these letters with 2 letters can be done in
\[\Rightarrow \dfrac{3!}{2!}\]
\[\Rightarrow 3\text{ ways}\]
Now, the 3-letter words that can be formed with two letters same is given by
\[\Rightarrow 2\times 6\times 3\]
Now, on further simplification we get,
\[\Rightarrow 36\]
Thus, there are 36 3- letter words with 2 letters same
Now, we need to find the number of letters with three letters same
Here, A is the only letter that is repeated thrice
So, the word can be AAA which has only 1 way
Now, the total number of 3-letter words that can be formed from the given word are
\[\Rightarrow 210+36+1\]
Now, on simplifying it further we get,
\[\Rightarrow 247\]
Hence, the correct option is (c).
Note:
It is important to note that we need to consider the words having 2 letters the same and three letters the same with the words having all letters different because these all satisfy the given condition. Here, neglecting any of the cases gives the incorrect option.
It is also to be noted that in the case that we considered having 2 letters to be the same after choosing the letters we need to arrange them because we are finding the words that are possible.
Complete step-by-step answer:
Now, from the given word SAHARANPUR in the question we have
S, A, A, A, H, R, R, N, P, U
Now, we need to find the 3 letter words that can be formed using these letters
Let us first consider the case that all the 3 letters to be different
Now, we have to arrange the 7 different letters in 3 places
As we already know that arrangement of n things in r places can be done using the permutations given by the formula
\[{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}\]
Now, on comparing with the formula we have
\[n=7,r=3\]
Now, on substituting the respective values we get,
\[\Rightarrow {}^{7}{{P}_{3}}\]
Now, this can be further written as
\[\Rightarrow \dfrac{7!}{\left( 7-3 \right)!}\]
Now, on further simplification we get,
\[\begin{align}
& \Rightarrow 7\times 6\times 5 \\
& \Rightarrow 210 \\
\end{align}\]
Thus, 210 3- letter words can be formed with all letters being different
Now, let us find the number of 3-letter words having two letters same
Here, the possible two same letters can be either A or R
As we already know that selection can be done using the combinations given by the formula
\[{}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}\]
Now, we can select either of A or R in
\[\Rightarrow {}^{2}{{C}_{1}}=\dfrac{2!}{1!1!}\]
That means in 2 ways
Now, the next letter can be any of the remaining 6 letters which can be done in
\[\Rightarrow 6\text{ ways}\]
Now, the arrangement of these letters with 2 letters can be done in
\[\Rightarrow \dfrac{3!}{2!}\]
\[\Rightarrow 3\text{ ways}\]
Now, the 3-letter words that can be formed with two letters same is given by
\[\Rightarrow 2\times 6\times 3\]
Now, on further simplification we get,
\[\Rightarrow 36\]
Thus, there are 36 3- letter words with 2 letters same
Now, we need to find the number of letters with three letters same
Here, A is the only letter that is repeated thrice
So, the word can be AAA which has only 1 way
Now, the total number of 3-letter words that can be formed from the given word are
\[\Rightarrow 210+36+1\]
Now, on simplifying it further we get,
\[\Rightarrow 247\]
Hence, the correct option is (c).
Note:
It is important to note that we need to consider the words having 2 letters the same and three letters the same with the words having all letters different because these all satisfy the given condition. Here, neglecting any of the cases gives the incorrect option.
It is also to be noted that in the case that we considered having 2 letters to be the same after choosing the letters we need to arrange them because we are finding the words that are possible.
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