Question

# The number of 5 letter words formed using letters of word “CALCULUS” is $(a){\text{ 280}} \\ (b){\text{ 15}} \\ (c){\text{ 1110}} \\ (d){\text{ 56}} \\$

Hint – In this question we have to find the number of ways of forming a 5 letter word out of the given alphabets of the word “CALCULUS”. Make different cases possible of making a 5 letter word as all the alphabets of the given words are not unique, so there will be cases where all 5 letters will be different or 2 will be the same only 3 will be different. Try to think of all possible scenarios using the above explained concept.

Complete step-by-step solution -

Given word
CALCULUS
Now we have to make a five letter word from this given word.
So, different words in the word calculus are A, C, C, L, L, S, U, U as we see there are five different words and C, L, U are coming at times.
Case 1 – when all the five different words are chosen.
So, the number of ways to form a five letter word is ${}^5{C_5} \times 5!$.
Where ${}^5{C_5}$ is selecting 5 words from the given five words and 5! Is the arrangement of these words.

Case 2 – when 2 same and 3 different words are chosen.
So, number of ways to form a five letter word is ${}^3{C_1} \times {}^4{C_3} \times \dfrac{{5!}}{{2!}}$ .
Where ${}^3{C_1}$ is selecting one pair from 3 pairs (LL, CC, UU), ${}^4{C_3}$ is selecting three words from the remaining four words and $\dfrac{{5!}}{{2!}}$ is the arrangement of these words, in which one word is repeated so we have to divide by 2!.

Case 3 – when 2 same, 2 same and 1 different words are chosen.
So, the number of ways to form a five letter word is ${}^3{C_2} \times {}^3{C_1} \times \dfrac{{5!}}{{2! \times 2!}}$ .
Where ${}^3{C_2}$ is selecting two pairs from 3 pairs (LL, CC, UU) , ${}^3{C_1}$ is selecting one word from the remaining three words and $\dfrac{{5!}}{{2! \times 2!}}$ is the arrangement of these words, in which two words are repeated so we have to divide by $2! \times 2!$.
So, the total no of ways to form a five letter word is,
$\Rightarrow {}^5{C_5} \times 5! + {}^3{C_1} \times {}^4{C_3} \times \dfrac{{5!}}{{2!}} + {}^3{C_2} \times {}^3{C_1} \times \dfrac{{5!}}{{2! \times 2!}}$
Now as we know ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ so, use this property in above equation we have,
$\Rightarrow \dfrac{{5!}}{{5!\left( {5 - 5} \right)!}} \times 5! + \dfrac{{3!}}{{1!\left( {3 - 1} \right)!}} \times \dfrac{{4!}}{{3!\left( {4 - 3} \right)!}} \times \dfrac{{5!}}{{2!}} + \dfrac{{3!}}{{2!\left( {3 - 2} \right)!}} \times \dfrac{{3!}}{{1!\left( {3 - 1} \right)!}} \times \dfrac{{5!}}{{2! \times 2!}}$
Now simplify the above equation we have,
$\Rightarrow 1 \times 5! + 3 \times 4 \times \dfrac{{5!}}{{2!}} + 3 \times 3 \times \dfrac{{5!}}{{2! \times 2!}}$
$\Rightarrow 120 + 12 \times \dfrac{{120}}{2} + 9 \times \dfrac{{120}}{4}$
$\Rightarrow 120 + 720 + 270$
$\Rightarrow 1110$
So, the total number of ways to form a five letter word is 1110.
Hence option (c) is correct.

Note – Whenever we face such types of problems the key point is simply to figure out all the possible scenarios which can arise while formation of the word. Permutation and combination has different meanings as combination resembles the ways of selecting whereas permutation resembles the ways of rearranging.