Question

Determine the number of arrangements of letters of the word ALGORITHM if

(a) vowels are always together.

(b) no two vowels are together.

(c) consonants are at even positions.

(d) O is the first and T is the last letter.

Answer 1

There are 9 letters in the word ALGORITHM.

(a) When vowels are always together. There are 3 vowels in the word ALGORITHM

(i.e., A, I, O). Let us consider these 3 vowels as one unit. This unit with 6 other letters is to be arranged.

∴ The number of arrangement = ⁷P₇ = 7! = 5040

3 vowels can be arranged among themselves in ³P₃ = 3! = 6 ways.

∴ Required number of arrangements = 7! × 3!

= 5040 × 6

= 30240

(b) When no two vowels are together. There are 6 consonants in the word ALGORITHM, they can be arranged among themselves in

⁶P₆ = 6! = 720 ways.

Let consonants be denoted by C.

_C _C_ C _C_C_C

There are 7 places marked by ‘_’ in which 3 vowels can be arranged.

∴ Vowels can be arranged in ⁷P₃ =

∴ Required number of arrangements = 720 × 210 = 151200

(c) When consonants are at even positions. There are 4 even places and 6 consonants in the word ALGORITHM. ∴ 6 consonants can be arranged at 4 even positions in 

⁶P₄ =\(\frac{6!}{(6-4)!}=\frac{6\times5\times4\times3\times2}{2!}\) = 360 ways.

Remaining 5 letters (3 vowels and 2 consonants) can be arranged in odd position in P = 5! = 120 ways.

∴ Required number of arrangements = 360 × 120 = 43200

(d) When O is the first and T is the last letter. All the letters of the word ALGORITHM are to be arranged among themselves such that arrangement begins with O and ends with T.

∴ Position of O and T are fixed.

∴ Other 7 letters can be arranged between O and T among themselves in ⁷P₇ = 7! = 5040 ways.

∴ Required number of arrangements = 5040