
Number of 4 letters words that can be formed with the letters of the word IIT JEE is:
A) 42
B) 82
C) 102
D) 142
Answer
545.1k+ views
Hint:
Here we will find the number of 4 letters words that can be formed using the Permutation and Combination concept. First, we will group up all the letters, taking the same letter together and different letters together. Then we will find the number of ways a word can be formed if we take all different letters. Next, we will find the number of ways a word can be formed if we take two different and two similar letters. Then we will find the number of ways if we take similar letters. Finally, we will add all the number of ways to get our desired answer.
Formula used:
\[n! = n \times \left( {n - 1} \right)\left( {n - 2} \right)........2 \times 1\], where \[n\] is the given number.
Complete step by step solution:
We have to find the number of words formed by using the letter of the word IIT JEE.
First, we will group the letters by taking similar letters together as follows:
(II) TJ (EE)
Here I letter is coming two times and E letter is also coming two times rest there is one T letter and one J letter.
Now, starting with a word having all letters different the number of ways to do that is,
The four different letters are (ITJE)
Number of ways for arranging four different letters \[ = 4! = 4 \times 3 \times 2 \times 1 = 24\]……\[\left( 1 \right)\]
Next, if there are two different and two similar letters in the word such as,
IITJ, EETJ, IITE, EETI, IIJE, EEJI
So for each word, there is \[4!\] ways but as two letters are repeating so we will divide it by \[2!\] as follows:
Number of ways two different and two similar letters \[ = \left( {4! \div 2!} \right) \times 6\]
Evaluating the factorial, we get
Number of ways of arranging two different and two similar letters \[ = \left( {\dfrac{{4 \times 3 \times 2 \times 2 \times 1}}{{2 \times 1}}} \right) \times 6\]
Simplifying the expression, we get
Number of ways of arranging two different and two similar letters \[ = 72\]….\[\left( 2 \right)\]
Lastly, if there are two similar and two similar letters in the word such as,
IIEE
So there is \[4!\] ways to get a number but as two letters are repeated so we will divide the number of ways by \[2!\]\[2!\] as follows.
Number of ways of arranging similar letters \[ = \left( {4!} \right) \div \left( {2! \times 2!} \right)\]
Computing the factorial, we get
Number of ways of arranging similar letters \[ = \dfrac{{4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 2 \times 1}} = 6\]……. \[\left( 3 \right)\]
Now adding equation \[\left( 1 \right)\], \[\left( 2 \right)\] and \[\left( 3 \right)\], we get
Total number of ways \[ = 24 + 72 + 6 = 102\]
Therefore, we can form 102 four-letters words from IIT JEE.
Hence, option (C) is correct.
Note:
Factorial of a number is the product of the entire positive number less than or equal to the number given. Factorials are used to calculate the permutation and combination of different letters or words to form something new. As there are \[n\] different ways to arrange any objects. The value of \[0!\] is 1, according to the convention of an empty product.
Here we will find the number of 4 letters words that can be formed using the Permutation and Combination concept. First, we will group up all the letters, taking the same letter together and different letters together. Then we will find the number of ways a word can be formed if we take all different letters. Next, we will find the number of ways a word can be formed if we take two different and two similar letters. Then we will find the number of ways if we take similar letters. Finally, we will add all the number of ways to get our desired answer.
Formula used:
\[n! = n \times \left( {n - 1} \right)\left( {n - 2} \right)........2 \times 1\], where \[n\] is the given number.
Complete step by step solution:
We have to find the number of words formed by using the letter of the word IIT JEE.
First, we will group the letters by taking similar letters together as follows:
(II) TJ (EE)
Here I letter is coming two times and E letter is also coming two times rest there is one T letter and one J letter.
Now, starting with a word having all letters different the number of ways to do that is,
The four different letters are (ITJE)
Number of ways for arranging four different letters \[ = 4! = 4 \times 3 \times 2 \times 1 = 24\]……\[\left( 1 \right)\]
Next, if there are two different and two similar letters in the word such as,
IITJ, EETJ, IITE, EETI, IIJE, EEJI
So for each word, there is \[4!\] ways but as two letters are repeating so we will divide it by \[2!\] as follows:
Number of ways two different and two similar letters \[ = \left( {4! \div 2!} \right) \times 6\]
Evaluating the factorial, we get
Number of ways of arranging two different and two similar letters \[ = \left( {\dfrac{{4 \times 3 \times 2 \times 2 \times 1}}{{2 \times 1}}} \right) \times 6\]
Simplifying the expression, we get
Number of ways of arranging two different and two similar letters \[ = 72\]….\[\left( 2 \right)\]
Lastly, if there are two similar and two similar letters in the word such as,
IIEE
So there is \[4!\] ways to get a number but as two letters are repeated so we will divide the number of ways by \[2!\]\[2!\] as follows.
Number of ways of arranging similar letters \[ = \left( {4!} \right) \div \left( {2! \times 2!} \right)\]
Computing the factorial, we get
Number of ways of arranging similar letters \[ = \dfrac{{4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 2 \times 1}} = 6\]……. \[\left( 3 \right)\]
Now adding equation \[\left( 1 \right)\], \[\left( 2 \right)\] and \[\left( 3 \right)\], we get
Total number of ways \[ = 24 + 72 + 6 = 102\]
Therefore, we can form 102 four-letters words from IIT JEE.
Hence, option (C) is correct.
Note:
Factorial of a number is the product of the entire positive number less than or equal to the number given. Factorials are used to calculate the permutation and combination of different letters or words to form something new. As there are \[n\] different ways to arrange any objects. The value of \[0!\] is 1, according to the convention of an empty product.
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