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Question: Consider a \(10\) letter word W= CURRICULUM. Number of ways in which \(5\)letter words can be formed...

Consider a 1010 letter word W= CURRICULUM. Number of ways in which 55letter words can be formed using the letter from the word “W” if each 55lettered word has exactly 33different letters, is
A) 360
B) 560
C) 610
D) 720

Explanation

Solution

To solve this question we need to have the knowledge of permutation and combination. The first step is to calculate the number of times each letter has occurred to form the word. Then as per the condition write all the cases required to form a word which question demands for.

Complete step by step answer:
The question asks us the number of ways in which a five letter word could be made with exactly three different letters from the word CURRICULUM. To solve this the first step will be to divide the problem in cases for making the calculation easier.
On analysing we see that in letter CURRICULUM, “C” occurs twice, “U” occurs thrice , “R” occurs twice , “I” occurs once , “L” occurs once , “M” occurs once. The cases that will be formed are:
Case1: For exactly three different letters all the “U” will be used with any two letters which are different like (“C”,”I”,“I”, “L”, “M”). Number of ways to select these letters are:
⇒5C2=5!2!(5−2)!\Rightarrow {}^{5}{{C}_{2}}=\dfrac{5!}{2!\left( 5-2 \right)!}
⇒5C2=5!2!(3)!\Rightarrow {}^{5}{{C}_{2}}=\dfrac{5!}{2!\left( 3 \right)!}
⇒5C2=5×4×3!2!(3)!\Rightarrow {}^{5}{{C}_{2}}=\dfrac{5\times 4\times 3!}{2!\left( 3 \right)!}
⇒5C2=10\Rightarrow {}^{5}{{C}_{2}}=10
The number of arrangement is =5!3!×10=200\dfrac{5!}{3!}\times 10=200
Case 2: Considering 2”C” and 2”R” to form a word of five letter. Number of ways to select these letters are:
⇒4C1=4!1!(4−1)!\Rightarrow {}^{4}{{C}_{1}}=\dfrac{4!}{1!\left( 4-1 \right)!}
⇒4C1=4!1!(3)!\Rightarrow {}^{4}{{C}_{1}}=\dfrac{4!}{1!\left( 3 \right)!}
⇒4C1=4\Rightarrow {}^{4}{{C}_{1}}=4
The number of arrangement is =5!2!2!×4=120\dfrac{5!}{2!2!}\times 4=120
Case 3: Considering 2”C” and 2”U” to form a word of five letter. Number of ways to select these letters are:
⇒4C1=4!1!(4−1)!\Rightarrow {}^{4}{{C}_{1}}=\dfrac{4!}{1!\left( 4-1 \right)!}
⇒4C1=4!1!(3)!\Rightarrow {}^{4}{{C}_{1}}=\dfrac{4!}{1!\left( 3 \right)!}
⇒4C1=4\Rightarrow {}^{4}{{C}_{1}}=4
The number of arrangement is =5!2!2!×4=120\dfrac{5!}{2!2!}\times 4=120
Case 4: Considering 2”U” and 2”R” to form a word of five letter. Number of ways to select these letters are:
⇒4C1=4!1!(4−1)!\Rightarrow {}^{4}{{C}_{1}}=\dfrac{4!}{1!\left( 4-1 \right)!}
⇒4C1=4!1!(3)!\Rightarrow {}^{4}{{C}_{1}}=\dfrac{4!}{1!\left( 3 \right)!}
⇒4C1=4\Rightarrow {}^{4}{{C}_{1}}=4
The number of arrangement is =5!2!2!×4=120\dfrac{5!}{2!2!}\times 4=120
Now we will be adding the arrangements of all the four cases. On doing this we get:
⇒200+120+120+120\Rightarrow 200+120+120+120
⇒560\Rightarrow 560 ways
∴\therefore Number of ways in which 55letter words can be formed using the letter from the word “W” if each 55 lettered word has exactly 33different letters, is Option B)560B)560.

So, the correct answer is “Option B”.

Note: To solve the question we need to remember the formula for the combination of a number. So the term nCr{}^{n}{{C}_{r}} is written as n!r!(n−r)!\dfrac{n!}{r!\left( n-r \right)!} . The term n!n! means that the number is being multiplied to the natural numbers below nn. This means n!=n×(n−1)×(n−2)........3×2×1n!=n\times \left( n-1 \right)\times \left( n-2 \right)........3\times 2\times 1.