Difficult Math Question #66 - Combinations

Total letter count = 8 (COMPUTER)

No of vowels = 3
Available positions for vowels = 4
No of free positions after vowel arrangement = 4-3 = 1

No of consonants= 5
Available positions for consonants= 5 (Inclusive vowel positions)

Lets say positions are=>1 2 3 4 5 6 7 8

3rd place is already fixed for M, leaves 4 consonants to be placed

Lets arrange consonants first==>

position number 1 can be arranged in 4 ways
position number 3 can be arranged in 1 ways
position number 5 can be arranged in 3 ways
position number 7 can be arranged in 2 ways

Leaves one vowel behind, which can be placed on available vowel position

remaining letters(3 vowels+1 consonant) can be arranged as 4! ways for available 4 positions

So the total posible arrangements = 4*1*3*2*4! = 576

800guy wrote:On how many ways can the letters of the word "COMPUTER" be arranged?
1. M must always occur at the third place
2. Vowels occupy the even positions.

Same logic as above, my answer is

4 * 3 * 1 * 2 * 3 * 1 * 2 * 1 = 144

OA:

For 1.
7*6*1*5*4*3*2*1=5,040

For 2.) I think It should be 4 * 720
there are 4 even positions to be filled by three even numbers.

in 5*3*4*2*3*1*2*1 It is assumed that Last even place is NOT filled by a vowel. There can be total 4 ways to do that.

Hence 4 * 720

How abt M's position???

Wont it affect?

lalitaroral wrote:How abt M's position???

Wont it affect?

don't know..that's the only OA provided.

800guy wrote:OA:

For 1.
7*6*1*5*4*3*2*1=5,040

For 2.) I think It should be 4 * 720
there are 4 even positions to be filled by three even numbers.

in 5*3*4*2*3*1*2*1 It is assumed that Last even place is NOT filled by a vowel. There can be total 4 ways to do that.

Hence 4 * 720

aren't 1) and 2) two conditions of the same problem?

Your answer seems like, they are two different questions.

Even if (1) and (2) are two independent problems...
the answers are different..

1) no of way the 7 letters can be arrange in 7 places available....7!


2) No. of even positions available for vowels...4
no. of ways 3 vowels can be arranged in 4 places = 4p3
now, no. of places left for remaining 5 letters = 5
no. of ways these 5 consonents can be arrnaged in 5 places left = 5!
i.e, = 4p3.5! = 480

** WHY IS THE LAST EVEN POSITION NOT AVAILABLE FOR FILLING A VOWEL , AS MENTIONED IN THE OA ?[/u][/i]