
Find the number of words that can be made with the letters of the word “GENIUS” if each word neither begins with G and nor ends with S
A.\[24\]
B.\[240\]
C.\[480\]
D.\[504\]
Answer
575.4k+ views
Hint:Here we count the number of letters in the word “GENIUS” and find a number of ways to write all the words where the word neither begins with G nor ends with N. We take 4 possibilities that are opposite to a word starting with G and ending with S and find the number of ways each position can be filled. Also, after assigning a letter to a position, always reduce the number of letters available by 1 for the next position.
Complete step-by-step answer:
We are given the word “GENIUS” which has a total number of letters as 6.
We have to make words from these six letters in such a way that word neither begins with G and nor ends with S.
We always assign a number of ways to choose letters for first and last position first and then from the remaining number of letters we find a number of ways to assign letters to other positions.
We take four possibilities.
First: Word starts with S and ends with G
Here we have fixed the first position as S and last position as G.
Number of ways to fill first position is 1 (only S)
Number of ways to fill last position is 1 (only G)
Now we have chosen 2 out of 6 letters. So for remaining positions we choose from 4 letters.
Number of ways to fill second position is 4
Number of ways to fill third position is 3
Number of ways to fill fourth position is 2
Number of ways to fill fifth position is 1
So, total number of words that can be formed in this way are: \[1 \times 4 \times 3 \times 2 \times 1 \times 1 = 24\]
Second: Word starts with S and G is somewhere in between
Here we have fixed the first position as S.
Number of ways to fill first position is 1 (only S)
Number of ways to fill last position is 4 (any letter other than G)
Now we have chosen 2 out of 6 letters. So for remaining positions we choose from 4 letters.
Number of ways to fill second position is 4
Number of ways to fill third position is 3
Number of ways to fill fourth position is 2
Number of ways to fill fifth position is 1
Number of ways to fill last position is 4 (any letter other than G)
So, total number of words that can be formed in this way are: \[1 \times 4 \times 3 \times 2 \times 1 \times 4 = 96\]
Third: Word ends with G and S is somewhere in between
Here we have fixed the last position as G.
Number of ways to fill first position is 4 (any letter other than G or S)
Number of ways to fill last position is 1 (only G)
Now we have chosen 2 out of 6 letters. So for remaining positions we choose from 4 letters.
Number of ways to fill second position is 4
Number of ways to fill third position is 3
Number of ways to fill fourth position is 2
Number of ways to fill fifth position is 1
So, total number of words that can be formed in this way are: \[4 \times 4 \times 3 \times 2 \times 1 \times 1 = 96\]
Fourth: Both S and G are somewhere in between but not on ends.
Here we have not fixed anything.
Number of ways to fill first position is 4 (any letter other than G or S)
Number of ways to fill last position is 3 (any letter other than G or S after choosing for first position)
Now we have chosen 2 out of 6 letters. So for remaining positions we choose from 4 letters.
Number of ways to fill second position is 4
Number of ways to fill third position is 3
Number of ways to fill fourth position is 2
Number of ways to fill fifth position is 1
Number of ways to fill last position is 3
So, total number of words that can be formed in this way are: \[4 \times 4 \times 3 \times 2 \times 1 \times 3 = 288\]
So, total number of word that can be formed will be the sum of words formed in each case
\[ \Rightarrow \]Total number of words \[ = 24 + 96 + 96 + 288\]
\[ \Rightarrow \]Total number of words \[ = 504\]
So, the correct answer is “Option D”.
Note:Students are likely to make mistakes in calculating the number of ways to fill positions when they don’t fill out first and last positions in the starting. Keep in mind to always assign a letter to the position on which condition is given.
Complete step-by-step answer:
We are given the word “GENIUS” which has a total number of letters as 6.
We have to make words from these six letters in such a way that word neither begins with G and nor ends with S.
We always assign a number of ways to choose letters for first and last position first and then from the remaining number of letters we find a number of ways to assign letters to other positions.
We take four possibilities.
First: Word starts with S and ends with G
Here we have fixed the first position as S and last position as G.
Number of ways to fill first position is 1 (only S)
Number of ways to fill last position is 1 (only G)
Now we have chosen 2 out of 6 letters. So for remaining positions we choose from 4 letters.
Number of ways to fill second position is 4
Number of ways to fill third position is 3
Number of ways to fill fourth position is 2
Number of ways to fill fifth position is 1
So, total number of words that can be formed in this way are: \[1 \times 4 \times 3 \times 2 \times 1 \times 1 = 24\]
Second: Word starts with S and G is somewhere in between
Here we have fixed the first position as S.
Number of ways to fill first position is 1 (only S)
Number of ways to fill last position is 4 (any letter other than G)
Now we have chosen 2 out of 6 letters. So for remaining positions we choose from 4 letters.
Number of ways to fill second position is 4
Number of ways to fill third position is 3
Number of ways to fill fourth position is 2
Number of ways to fill fifth position is 1
Number of ways to fill last position is 4 (any letter other than G)
So, total number of words that can be formed in this way are: \[1 \times 4 \times 3 \times 2 \times 1 \times 4 = 96\]
Third: Word ends with G and S is somewhere in between
Here we have fixed the last position as G.
Number of ways to fill first position is 4 (any letter other than G or S)
Number of ways to fill last position is 1 (only G)
Now we have chosen 2 out of 6 letters. So for remaining positions we choose from 4 letters.
Number of ways to fill second position is 4
Number of ways to fill third position is 3
Number of ways to fill fourth position is 2
Number of ways to fill fifth position is 1
So, total number of words that can be formed in this way are: \[4 \times 4 \times 3 \times 2 \times 1 \times 1 = 96\]
Fourth: Both S and G are somewhere in between but not on ends.
Here we have not fixed anything.
Number of ways to fill first position is 4 (any letter other than G or S)
Number of ways to fill last position is 3 (any letter other than G or S after choosing for first position)
Now we have chosen 2 out of 6 letters. So for remaining positions we choose from 4 letters.
Number of ways to fill second position is 4
Number of ways to fill third position is 3
Number of ways to fill fourth position is 2
Number of ways to fill fifth position is 1
Number of ways to fill last position is 3
So, total number of words that can be formed in this way are: \[4 \times 4 \times 3 \times 2 \times 1 \times 3 = 288\]
So, total number of word that can be formed will be the sum of words formed in each case
\[ \Rightarrow \]Total number of words \[ = 24 + 96 + 96 + 288\]
\[ \Rightarrow \]Total number of words \[ = 504\]
So, the correct answer is “Option D”.
Note:Students are likely to make mistakes in calculating the number of ways to fill positions when they don’t fill out first and last positions in the starting. Keep in mind to always assign a letter to the position on which condition is given.
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