The number of five-letter words formed using the letters of the word “CALCULUS” is
(a) 280
(b) 15
(c) 1110
(d) 56
Answer
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Hint:
Here, we need to find the number of five-letter words that can be formed by using the letters of the word “CALCULUS”. We will use different cases and add the results together to find the required number of words. The different cases are – no letter is repeated, C is repeated, U is repeated, L is repeated, C and L are repeated, C and U are repeated, and L and U are repeated.
Formula Used:
We will use the following formulas:
The number of permutations in which a set of \[n\] objects can be arranged in \[r\] places is given by \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\], where no object is repeated.
The number of permutations to arrange \[n\] objects is given by \[\dfrac{{n!}}{{{r_1}!{r_2}! \ldots {r_n}!}}\], where an object appears \[{r_1}\] times, another object repeats \[{r_2}\], and so on.
Complete step by step solution:
The number of letters in the word CALCULUS are 8, where C, L, and U come twice.
The letters are to be arranged in 5 places.
There are 5 distinct letters in the word CALCULUS, that is C, A, L, U, S.
Thus, we can find the answer using seven cases.
Case 1: The letters C, L, U are not repeated.
We have 5 distinct letters to be placed in 5 spaces, where no letter is being repeated.
Substituting \[n = 5\] and \[r = 5\] in the formula \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\], we get
\[{}^5{P_5} = \dfrac{{5!}}{{\left( {5 - 5} \right)!}} = \dfrac{{5!}}{{0!}} = 5! = 120\]
Therefore, the number of five-letter words that can be formed when C, L, U are not repeated is 120 words.
Case 2: The letter C is repeated in the 5 places, but letters L and U are not repeated.
In the 5 places, 2 places will be taken by the two C’s, and the remaining 3 places will be taken by any of the remaining 4 letters, that is A, L, U, S.
The number of ways in which this is possible can be found by using combinations.
Therefore, we get
Number of five-letter words where C is repeated, but L, U are not repeated (order not important) \[ = {}^2{C_2} \times {}^4{C_3}\]
Since the order matters in the number of words we need to find, we will find the order in which the 5 letters (chosen in \[{}^2{C_2} \times {}^4{C_3}\] ways) can be placed in the 5 places, where C is repeated.
This can be found by using the formula \[\dfrac{{n!}}{{{r_1}!{r_2}! \ldots {r_n}!}}\].
Thus, the five chosen letters can be ordered in \[\dfrac{{5!}}{{2!}}\] ways.
Therefore, we get the number of five-letter words where the C is repeated, but L and U are not repeated, is given by \[{}^2{C_2} \times {}^4{C_3} \times \dfrac{{5!}}{{2!}}\] ways.
Here, \[{}^2{C_2} \times {}^4{C_3}\] is the number of ways of choosing the letters to be placed within the 4 places, and \[\dfrac{{5!}}{{2!}}\] is the number of ways in which the chosen 4 letters can be ordered.
Simplifying the expression, we get
Number of five-letter words where C is repeated, but L, U are not repeated
\[ = 1 \times 4 \times \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}} = 240\]
Therefore, the number of five-letter words where the letter C is repeated, but L, U are not repeated, is 240 words.
Case 3: The letter L is repeated in the 5 places, but letters C and U are not repeated.
In the 5 places, 2 places will be taken by the two L’s, and the remaining 3 places will be taken by any of the remaining 4 letters, that is A, C, U, S.
Therefore, we get
Number of five-letter words where L is repeated, but C, U are not repeated (order not important) \[ = {}^2{C_2} \times {}^4{C_3}\]
Since the order matters in the number of words we need to find, we will find the order in which the 5 letters (chosen in \[{}^2{C_2} \times {}^4{C_3}\] ways) can be placed in the 5 places, where L is repeated.
Thus, the five chosen letters can be ordered in \[\dfrac{{5!}}{{2!}}\] ways.
Therefore, we get the number of five-letter words where the L is repeated, but C and U are not repeated, is given by \[{}^2{C_2} \times {}^4{C_3} \times \dfrac{{5!}}{{2!}}\] ways.
Simplifying the expression, we get
Number of five-letter words where L is repeated, but C, U are not repeated
\[ = 1 \times 4 \times \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}} = 240\]
Therefore, the number of five-letter words where the letter L is repeated, but C, U are not repeated, is 240 words.
Case 4: The letter U is repeated in the 5 places, but letters C and L are not repeated.
In the 5 places, 2 places will be taken by the two U’s, and the remaining 3 places will be taken by any of the remaining 4 letters, that is A, C, L, S.
Therefore, we get
Number of five-letter words where U is repeated, but C, L are not repeated (order not important) \[ = {}^2{C_2} \times {}^4{C_3}\]
Since the order matters in the number of words we need to find, we will find the order in which the 5 letters (chosen in \[{}^2{C_2} \times {}^4{C_3}\] ways) can be placed in the 5 places, where U is repeated.
Thus, the five chosen letters can be ordered in \[\dfrac{{5!}}{{2!}}\] ways.
Therefore, we get the number of five-letter words where the U is repeated, but C and L are not repeated, is given by \[{}^2{C_2} \times {}^4{C_3} \times \dfrac{{5!}}{{2!}}\] ways.
Simplifying the expression, we get
Number of five-letter words where U is repeated, but C, L are not repeated
\[ = 1 \times 4 \times \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}} = 240\]
Therefore, the number of five-letter words where the letter U is repeated, but C, L are not repeated, is 240 words.
Case 5: The letters C and L are repeated in the 5 places, but letter U is not repeated.
In the 5 places, 2 places will be taken by the two C’s, 2 places will be taken by the two L’s and the remaining 1 place will be taken by any of the remaining 3 letters, that is A, U, S.
Therefore, we get
Number of five-letter words where C and L are repeated, but U is not repeated (order not important) \[ = {}^2{C_2} \times {}^2{C_2} \times {}^3{C_1}\]
Since the order matters in the number of words we need to find, we will find the order in which the 5 letters (chosen in \[{}^2{C_2} \times {}^2{C_2} \times {}^3{C_1}\] ways) can be placed in the 5 places, where C, L are repeated.
Thus, the five chosen letters can be ordered in \[\dfrac{{5!}}{{2!2!}}\] ways.
Therefore, we get the number of five-letter words where the C, L are repeated, but U is not repeated, is given by \[{}^2{C_2} \times {}^2{C_2} \times {}^3{C_1} \times \dfrac{{5!}}{{2!2!}}\] ways.
Simplifying the expression, we get
Number of five-letter words where C and L are repeated, but U is not repeated
\[ = 1 \times 1 \times 3 \times \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 2 \times 1}} = 90\]
Therefore, the number of five-letter words where the letter C and L are repeated, but U is not repeated, is 90 words.
Case 6: The letters C and U are repeated in the 5 places, but letter L is not repeated.
In the 5 places, 2 places will be taken by the two C’s, 2 places will be taken by the two U’s and the remaining 1 place will be taken by any of the remaining 3 letters, that is A, L, S.
Therefore, we get
Number of five-letter words where C and U are repeated, but L is not repeated (order not important) \[ = {}^2{C_2} \times {}^2{C_2} \times {}^3{C_1}\]
Since the order matters in the number of words we need to find, we will find the order in which the 5 letters (chosen in \[{}^2{C_2} \times {}^2{C_2} \times {}^3{C_1}\] ways) can be placed in the 5 places, where C, U are repeated.
Thus, the five chosen letters can be ordered in \[\dfrac{{5!}}{{2!2!}}\] ways.
Therefore, we get the number of five-letter words where the C, U are repeated, but L is not repeated, is given by \[{}^2{C_2} \times {}^2{C_2} \times {}^3{C_1} \times \dfrac{{5!}}{{2!2!}}\] ways.
Simplifying the expression, we get
Number of five-letter words where C and U are repeated, but L is not repeated
\[ = 1 \times 1 \times 3 \times \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 2 \times 1}} = 90\]
Therefore, the number of five-letter words where the letter C and U are repeated, but L is not repeated, is 90 words.
Case 7: The letters U and L are repeated in the 5 places, but letter C is not repeated.
In the 5 places, 2 places will be taken by the two U’s, 2 places will be taken by the two L’s and the remaining 1 place will be taken by any of the remaining 3 letters, that is A, C, S.
Therefore, we get
Number of five-letter words where U and L are repeated, but C is not repeated (order not important) \[ = {}^2{C_2} \times {}^2{C_2} \times {}^3{C_1}\]
Since the order matters in the number of words we need to find, we will find the order in which the 5 letters (chosen in \[{}^2{C_2} \times {}^2{C_2} \times {}^3{C_1}\] ways) can be placed in the 5 places, where U, L are repeated.
Thus, the five chosen letters can be ordered in \[\dfrac{{5!}}{{2!2!}}\] ways.
Therefore, we get the number of five-letter words where the U, L are repeated, but C is not repeated, is given by \[{}^2{C_2} \times {}^2{C_2} \times {}^3{C_1} \times \dfrac{{5!}}{{2!2!}}\] ways.
Simplifying the expression, we get
Number of five-letter words where U and L are repeated, but C is not repeated
\[ = 1 \times 1 \times 3 \times \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 2 \times 1}} = 90\]
Therefore, the number of five-letter words where the letter U and L are repeated, but C is not repeated, is 90 words.
Finally, we will calculate the number of five-letter words that can be formed using the letters of the word “CALCULUS”.
The number of five letter words that can be formed is the sum of all the possible cases.
Thus, we get the number of five-letter words that can be formed using the letters of the word “CALCULUS” as \[120 + 240 + 240 + 240 + 90 + 90 + 90 = 1110\] words.
Therefore, 1110 five-letter words can be formed using the letters of the word “CALCULUS”.
Thus, the correct option is option (c).
Note:
We used combinations in the solution to find the number of ways where order is not important. The number of combinations in which a set of \[n\] objects can be arranged in \[r\] places is given by \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]. Therefore, the number of ways in which the 2 C’s can be placed in 2 places is \[{}^2{C_2}\], and the number of ways to place the remaining 4 letters in the 3 places is \[{}^4{C_3}\]. By multiplying these, we get the number of ways to place the 5 letters in the 5 places, such that the letter C comes twice, and order of letters does not matter.
Here, we need to find the number of five-letter words that can be formed by using the letters of the word “CALCULUS”. We will use different cases and add the results together to find the required number of words. The different cases are – no letter is repeated, C is repeated, U is repeated, L is repeated, C and L are repeated, C and U are repeated, and L and U are repeated.
Formula Used:
We will use the following formulas:
The number of permutations in which a set of \[n\] objects can be arranged in \[r\] places is given by \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\], where no object is repeated.
The number of permutations to arrange \[n\] objects is given by \[\dfrac{{n!}}{{{r_1}!{r_2}! \ldots {r_n}!}}\], where an object appears \[{r_1}\] times, another object repeats \[{r_2}\], and so on.
Complete step by step solution:
The number of letters in the word CALCULUS are 8, where C, L, and U come twice.
The letters are to be arranged in 5 places.
There are 5 distinct letters in the word CALCULUS, that is C, A, L, U, S.
Thus, we can find the answer using seven cases.
Case 1: The letters C, L, U are not repeated.
We have 5 distinct letters to be placed in 5 spaces, where no letter is being repeated.
Substituting \[n = 5\] and \[r = 5\] in the formula \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\], we get
\[{}^5{P_5} = \dfrac{{5!}}{{\left( {5 - 5} \right)!}} = \dfrac{{5!}}{{0!}} = 5! = 120\]
Therefore, the number of five-letter words that can be formed when C, L, U are not repeated is 120 words.
Case 2: The letter C is repeated in the 5 places, but letters L and U are not repeated.
In the 5 places, 2 places will be taken by the two C’s, and the remaining 3 places will be taken by any of the remaining 4 letters, that is A, L, U, S.
The number of ways in which this is possible can be found by using combinations.
Therefore, we get
Number of five-letter words where C is repeated, but L, U are not repeated (order not important) \[ = {}^2{C_2} \times {}^4{C_3}\]
Since the order matters in the number of words we need to find, we will find the order in which the 5 letters (chosen in \[{}^2{C_2} \times {}^4{C_3}\] ways) can be placed in the 5 places, where C is repeated.
This can be found by using the formula \[\dfrac{{n!}}{{{r_1}!{r_2}! \ldots {r_n}!}}\].
Thus, the five chosen letters can be ordered in \[\dfrac{{5!}}{{2!}}\] ways.
Therefore, we get the number of five-letter words where the C is repeated, but L and U are not repeated, is given by \[{}^2{C_2} \times {}^4{C_3} \times \dfrac{{5!}}{{2!}}\] ways.
Here, \[{}^2{C_2} \times {}^4{C_3}\] is the number of ways of choosing the letters to be placed within the 4 places, and \[\dfrac{{5!}}{{2!}}\] is the number of ways in which the chosen 4 letters can be ordered.
Simplifying the expression, we get
Number of five-letter words where C is repeated, but L, U are not repeated
\[ = 1 \times 4 \times \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}} = 240\]
Therefore, the number of five-letter words where the letter C is repeated, but L, U are not repeated, is 240 words.
Case 3: The letter L is repeated in the 5 places, but letters C and U are not repeated.
In the 5 places, 2 places will be taken by the two L’s, and the remaining 3 places will be taken by any of the remaining 4 letters, that is A, C, U, S.
Therefore, we get
Number of five-letter words where L is repeated, but C, U are not repeated (order not important) \[ = {}^2{C_2} \times {}^4{C_3}\]
Since the order matters in the number of words we need to find, we will find the order in which the 5 letters (chosen in \[{}^2{C_2} \times {}^4{C_3}\] ways) can be placed in the 5 places, where L is repeated.
Thus, the five chosen letters can be ordered in \[\dfrac{{5!}}{{2!}}\] ways.
Therefore, we get the number of five-letter words where the L is repeated, but C and U are not repeated, is given by \[{}^2{C_2} \times {}^4{C_3} \times \dfrac{{5!}}{{2!}}\] ways.
Simplifying the expression, we get
Number of five-letter words where L is repeated, but C, U are not repeated
\[ = 1 \times 4 \times \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}} = 240\]
Therefore, the number of five-letter words where the letter L is repeated, but C, U are not repeated, is 240 words.
Case 4: The letter U is repeated in the 5 places, but letters C and L are not repeated.
In the 5 places, 2 places will be taken by the two U’s, and the remaining 3 places will be taken by any of the remaining 4 letters, that is A, C, L, S.
Therefore, we get
Number of five-letter words where U is repeated, but C, L are not repeated (order not important) \[ = {}^2{C_2} \times {}^4{C_3}\]
Since the order matters in the number of words we need to find, we will find the order in which the 5 letters (chosen in \[{}^2{C_2} \times {}^4{C_3}\] ways) can be placed in the 5 places, where U is repeated.
Thus, the five chosen letters can be ordered in \[\dfrac{{5!}}{{2!}}\] ways.
Therefore, we get the number of five-letter words where the U is repeated, but C and L are not repeated, is given by \[{}^2{C_2} \times {}^4{C_3} \times \dfrac{{5!}}{{2!}}\] ways.
Simplifying the expression, we get
Number of five-letter words where U is repeated, but C, L are not repeated
\[ = 1 \times 4 \times \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}} = 240\]
Therefore, the number of five-letter words where the letter U is repeated, but C, L are not repeated, is 240 words.
Case 5: The letters C and L are repeated in the 5 places, but letter U is not repeated.
In the 5 places, 2 places will be taken by the two C’s, 2 places will be taken by the two L’s and the remaining 1 place will be taken by any of the remaining 3 letters, that is A, U, S.
Therefore, we get
Number of five-letter words where C and L are repeated, but U is not repeated (order not important) \[ = {}^2{C_2} \times {}^2{C_2} \times {}^3{C_1}\]
Since the order matters in the number of words we need to find, we will find the order in which the 5 letters (chosen in \[{}^2{C_2} \times {}^2{C_2} \times {}^3{C_1}\] ways) can be placed in the 5 places, where C, L are repeated.
Thus, the five chosen letters can be ordered in \[\dfrac{{5!}}{{2!2!}}\] ways.
Therefore, we get the number of five-letter words where the C, L are repeated, but U is not repeated, is given by \[{}^2{C_2} \times {}^2{C_2} \times {}^3{C_1} \times \dfrac{{5!}}{{2!2!}}\] ways.
Simplifying the expression, we get
Number of five-letter words where C and L are repeated, but U is not repeated
\[ = 1 \times 1 \times 3 \times \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 2 \times 1}} = 90\]
Therefore, the number of five-letter words where the letter C and L are repeated, but U is not repeated, is 90 words.
Case 6: The letters C and U are repeated in the 5 places, but letter L is not repeated.
In the 5 places, 2 places will be taken by the two C’s, 2 places will be taken by the two U’s and the remaining 1 place will be taken by any of the remaining 3 letters, that is A, L, S.
Therefore, we get
Number of five-letter words where C and U are repeated, but L is not repeated (order not important) \[ = {}^2{C_2} \times {}^2{C_2} \times {}^3{C_1}\]
Since the order matters in the number of words we need to find, we will find the order in which the 5 letters (chosen in \[{}^2{C_2} \times {}^2{C_2} \times {}^3{C_1}\] ways) can be placed in the 5 places, where C, U are repeated.
Thus, the five chosen letters can be ordered in \[\dfrac{{5!}}{{2!2!}}\] ways.
Therefore, we get the number of five-letter words where the C, U are repeated, but L is not repeated, is given by \[{}^2{C_2} \times {}^2{C_2} \times {}^3{C_1} \times \dfrac{{5!}}{{2!2!}}\] ways.
Simplifying the expression, we get
Number of five-letter words where C and U are repeated, but L is not repeated
\[ = 1 \times 1 \times 3 \times \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 2 \times 1}} = 90\]
Therefore, the number of five-letter words where the letter C and U are repeated, but L is not repeated, is 90 words.
Case 7: The letters U and L are repeated in the 5 places, but letter C is not repeated.
In the 5 places, 2 places will be taken by the two U’s, 2 places will be taken by the two L’s and the remaining 1 place will be taken by any of the remaining 3 letters, that is A, C, S.
Therefore, we get
Number of five-letter words where U and L are repeated, but C is not repeated (order not important) \[ = {}^2{C_2} \times {}^2{C_2} \times {}^3{C_1}\]
Since the order matters in the number of words we need to find, we will find the order in which the 5 letters (chosen in \[{}^2{C_2} \times {}^2{C_2} \times {}^3{C_1}\] ways) can be placed in the 5 places, where U, L are repeated.
Thus, the five chosen letters can be ordered in \[\dfrac{{5!}}{{2!2!}}\] ways.
Therefore, we get the number of five-letter words where the U, L are repeated, but C is not repeated, is given by \[{}^2{C_2} \times {}^2{C_2} \times {}^3{C_1} \times \dfrac{{5!}}{{2!2!}}\] ways.
Simplifying the expression, we get
Number of five-letter words where U and L are repeated, but C is not repeated
\[ = 1 \times 1 \times 3 \times \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 2 \times 1}} = 90\]
Therefore, the number of five-letter words where the letter U and L are repeated, but C is not repeated, is 90 words.
Finally, we will calculate the number of five-letter words that can be formed using the letters of the word “CALCULUS”.
The number of five letter words that can be formed is the sum of all the possible cases.
Thus, we get the number of five-letter words that can be formed using the letters of the word “CALCULUS” as \[120 + 240 + 240 + 240 + 90 + 90 + 90 = 1110\] words.
Therefore, 1110 five-letter words can be formed using the letters of the word “CALCULUS”.
Thus, the correct option is option (c).
Note:
We used combinations in the solution to find the number of ways where order is not important. The number of combinations in which a set of \[n\] objects can be arranged in \[r\] places is given by \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]. Therefore, the number of ways in which the 2 C’s can be placed in 2 places is \[{}^2{C_2}\], and the number of ways to place the remaining 4 letters in the 3 places is \[{}^4{C_3}\]. By multiplying these, we get the number of ways to place the 5 letters in the 5 places, such that the letter C comes twice, and order of letters does not matter.
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