Question

The number of 5 letter words formed using letters of word "CALCULUS" is
(A) 280
(B) 15
(C) 1110
(D) 56

Answer 1

According to the given statement it is clear that the problem is based on general permutation and combination. Here, we are using the basic formulas to calculate the number of ways, 2 same and 3 different letters, and 2 same, 2 same, and 1 different letter. Then by adding all the calculations, we can easily calculate the number of 5 letter words using the word CALCULUS.

Formula used:
Here, we use the formula of combination that is $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$

Complete step-by-step answer:
As there are 8 letters in the word CALCULUS.
In which the letters are, C (2 times), L (2 times), U (2 times), A and S.
Here we are taking 5 different words that are CLUAS which means all 5 get selected and their arrangements.
Therefore, we get $^5{C_2} \times 5!$
Opening $^5{C_2}$ using the formula $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ .As we can see n is 5 and r is 2.
So we will substitute in the formula to get,
$\Rightarrow \dfrac{{5!}}{{2!\left( {5 - 2} \right)!}} \times 5!$
$\Rightarrow \dfrac{{5!}}{{2!\left( 3 \right)!}} \times 5!$
On opening and cancelling the factorials we get,
$\Rightarrow \dfrac{{5!}}{{2!\left( 3 \right)!}} \times 3! \times 2!$ which is equal to $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$
Therefore, total number of ways = 120
Now we will select 2 same and 3 different words. For example: CC LUA that is from 3 same pairs we choose out 1 pair and from 4 different words we choose 3 combinations and their arrangements too.
Therefore, we get $^3{C_1}{ \times ^4}{C_3} \times \dfrac{{5!}}{{2!}}$
By using the above formula of combination.
$\Rightarrow \dfrac{{3!}}{{1!\left( {3 - 1} \right)!}} \times \dfrac{{4!}}{{3!\left( {4 - 3} \right)!}} \times \dfrac{{5!}}{{2!}}$
We can further solve it by cancelling and opening the factorials.
$\Rightarrow 720$
Now we will select 2 same, 2 same and 1 different words. For example: CC LL U that is from 3 same pairs we choose out 2 pairs and from 3 different words we choose 1 combination and their arrangements too.
Therefore, we get $^3{C_2}{ \times ^3}{C_1} \times \dfrac{{5!}}{{2!2!}}$
By using the above formula of combination.
$\Rightarrow \dfrac{{3!}}{{2!\left( {3 - 2} \right)!}} \times \dfrac{{3!}}{{1!\left( {3 - 1} \right)!}} \times \dfrac{{5!}}{{2!2!}}$
We can further solve it by cancelling and opening the factorials.
$\Rightarrow 270$
Hence, total words formed by using the letters of word CALCULUS is $120 + 720 + 270$
We get, $1110$ words
So, option (C) $1110$ is correct.

Note: To solve these types of questions, we need to find the letters which are repeating in the given word. As we know the letters are repeated twice in the given questions so do not forget to divide with $2!$ .