Permutations and Combinations Question?

Question

Find the number of ways you can arrange the letters of the word "MALENKOV" such that no 2 vowels are together.

Answer 1

First, how many ways can we arrange the three vowels among the eight letters? This is equivalent to counting the number of eight-bit binary numbers with exactly three ones and no two ones adjacent. (I'd use Cs and Vs for consonants and vowels, but I want to avoid any possible symbolic confusion--V is one of the letters we're using.)

These can be counted systematically. This method starts with all the ones in the rightmost positions possible and progresses from one arrangement to the next by shifting to the left the rightmost one that can be shifted, repositioning all ones to the right of it to their rightmost positions again:

00010101

00100101

00101001

00101010

01000101

01001001

01001010

01010001

01010010

01010100

10000101

10001001

10001010

10010001

10010010

10010100

10100001

10100010

10100100

10101000

I got 20 arrangements of the vowels, interspersed among the consonants. For each of these, the vowels can be order 3! different ways and the consonants 5! different ways, so the total number is

20 * 3! * 5! = 20 * 6 * 120 = 120² = 14,400

Answer 2

Answer is 20 * 3! * 6P3 = 20 * 6 * 120 = 17280 ways.

Procedure to solve :

First write the 3 vowels AEO.

Now, we have to place the remaing 5 letters such that no 3 of these vowels are together. Therefore put any 2 letters in 2 places [ (b/w A and E )and (b/w E and O) ] . To do this there are 5 * 4 = 20 ways. Now also arrange the AEO, total ways is 3! (3 factorial). Now arrange the remaining 3 letters in 6 positions, total ways 6P3.

Answer 3

First of all, reduce your problem to:

How many rearrangements of 0,0,0,0,0,1,1,1 have no adjacent 1s?

We use 0s for consonants and 1s for vowels. That lets us count something easier. There are 56 = 8C3 rearrangements.

In order to determine which of those have no adjacent 1s, you'll have to use the principle of inclusion/exclusion. Define the following:

P(≥1) = total arrangements

P(≥2) = arrangements with at least two consecutive vowels*

P(≥3) = arrangements with at least three consecutive vowels

P(=1) = what you want (no consecutive vowels)

P(=2) = arrangements with exactly two consecutive vowels**

P(=3) = arrangements with exactly three consecutive vowels

*counted in a specific way below, including some "over-counting"

**not used

What we know tells us:

P(≥1) = 56

We also know that since there are only 3 vowels:

P(≥3) = P(=3)

If we clump the 1s together, we can find that P(=3) = 6C1 = 6. You simply consider the six open "slots" - one gets the clump of 1s, the rest get 0s.

Likewise, if we clump two of the 1s, we can arrange that clump and the other 1 into the fold and get P(≥2) = 7×6 = 42. Same idea, of the seven open slots, one gets the clump of 1s (7 choices) the other gets the spare 1.

Now, the sieve principle tells us that in P(≥2) we also counted possibilities where the lone 1 and the clump of 1s were together - making the larger run of three 1s counted in P(=3).

Because of that, we simply do the following:

P(=1) = P(≥1) - P(≥2) + P(≥3) = 56 - 42 + 6 = 20

See:

http://en.wikipedia.org/wiki/Inclusion-e...

That compensates for "double counting." What happens is, if you take all of them and say "remove all the P(≥2) ones" you've actually counted the P(=3) ones twice, by removing them with the ones clumped like this [11]1 and like this 1[11]. So add back in the extra 6 that got counted double, and you get the answer.

Then it's just the simple matter of filling in the letters. We've determined where the vowels go (3 of them, all distinct) and where the consonants go (5 of them, all distinct) so your answer is:

total = 20 × 3! × 5! = 20 × 6 × 120 = 14,400

Answer 4

There may be a way to stick this into a formula and get an answer out, but I don't know such a formula, so here's how I would do it: Every vowel either precedes a consonant or is last.

How many such words are there where a vowel is last?

3C1 = 3 choices for which vowel goes last.

5P2 = 20 ways to then pair up consonants to follow the other two vowels.

5! = 120 ways to order the pieces.

3*20*120 = 7200 such words.

How many such words are there where a consonant is last?

5P3 = 60 ways to pair up consonants to follow the vowels.

5! = 120 ways to order the pieces.

60*120 = 7200 such words.

7200 + 7200 = 14400 total words such that no 2 vowels are together.