Question

Find the number of arrangements that can be made out of the letters of the word COMBINATION. In how many of these, vowels occur together?

Answer 1

Answer:

COMBINATION can be arranged in 4,989,600 different ways if it is eleven letters and only use each letter once. Assuming all vowels will be together 604,800 arrangements.

Explanation:

Requires work with permutations and factorial.

Factorial is written as '!'. Factorial is the multiplication of all it lower terms.

Eg 11! = 11×10×9×8×7×6×5×4×3×2×1

COMBINATION has eleven letters and as such can be arranged 11! ways. As it has repeating letters you divide by this repetitions.

As such it becomes 11!/2!.2!.2! . This equation equals 4,989,600

For the second part it should be treated as two parts . All the vowels are grouped together so there are effectively only 6 letters left.

This you would write as 8!/2!.2! (Lose a 2! as the repeating vowel is not counted. ) Then for the vowel,it is 5!/2!

You now multiply these together to get your answer of 604,800

Answer 2

Given ,

There are 11 letters in word COMBINATION , in which each O , I , and N appears 2 times

In word COMBINATION , vowels are A , I , and O

Let ,

A , 2 I's , 2 O's be the single word

Thus ,

Number of arrangement = 6! = 720 ways

Now , the vowels can be arranged in

$\sf \implies \frac{5!</p><p>}{2 !</p><p>\times 2!</p><p>} = \frac{5 \times 4 \times 3 \times 2}{2 \times 2} = 30 \: ways$

Therefore ,

No. of ways in which all vowels occur together will be

$\tt \implies 720 \times 30 = 21600 \: \: ways$