# Permutations question

1. Dec 20, 2012

### skg94

1. The problem statement, all variables and given/known data

1.How many 4 digit even numbers greater than 5000 can you form using the digits 0,1,2,3,5,6,8 and 9 without repetitions? How do i go about this?

b) how many of these numbers end in 0?

2. n_P_3=120

3. How many ways can all the letters of aloha be arranged if each arrangment must begin with a vowel and the consonants cannot be together?

2. Relevant equations

3. The attempt at a solution
1. im doing the cases, and i set it up as 1*7*6*2 + 3*7*6*3 which isnt right
answer is 420, b) is 120

2. n!/(n-3)! = 120 = n(n-1)(n-2)=120 doesnt make sense from here.

3. I was thinking 3*4*2*1 = 24 which isnt right either. answer is 18

2. Dec 20, 2012

### Vineeth T

Soln for 1:

a)If a number should be even its last digit should be even.
So the four digit number should end in 2,6,8,0.
Now consider two cases.
case I:The last digit is either 0 or 2(even numbers<5)
Now we should count how many such numbers are there.
We have four spaces _ _ _ _;
The last space can have two values(0,2).
The first space can have four values(5,6,8,9).
The second space can have 6 values (any of the eight numbers excluding the numbers used in the first and the last digit)
The third space can have 5 values (similar explanation)
From the principle of multiplication we can say that total number of numbers=2*4*6*5
=240
case II:The last digit is either 6 or 8(even numbers>5)
Similar to case I;
The last space can take two values(6,8)
The first space can take 3 values(any of 5,6,8,9 except the number used in last space since repetition not allowed)
The second and third spaces can take the 6&5 values respectively.(similar to case I)
So such numbers will be 2*3*6*5=180

b)The last digit is zero;
so the first digit can have 4 values(5,6,8,9)
the last digit can have 1 value(0)
the second & third digit can have 6&5 values respectively.
Hence such numbers=4*1*6*5
=120!!!

3. Dec 20, 2012

### Vineeth T

Soln for 2:

Split 120 into product of three factors which are in ascending or descending order.
The largest of the three will be n.

120=6*5*4
=n*n-1*n-2

By comparing n=6!!!

4. Dec 20, 2012

### pasmith

For 3:

I really don't see where the answer 18 comes from.

The possible arrangements of vowels and consonants are VCVCV and VVCVC. For the vowels, there are three possible places for the 'o' and the other two are then taken up by 'a's, so there are 3 essentially different arrangements. For the consonants there are 2 arrangements, so that the total is 3*2*2 = 12. If you distinguish between 'a's there are instead 3!*2*2 = 24 arrangements.

5. Dec 20, 2012

### Vineeth T

Soln for 3:

In this case the best way is to write all such words (I tried to explain as previous question but it took a lot of time to type so I left it)
Since in the question it is given that "All the letters of ALOHA"(so we should use a,l,o,h) we can have four letter words also.
a l o h a
a h o l a
a l a h o
a h a l o
a a l o h
a a h o l
a o l a h
a o h a l
o l a h a
o h a l a
o a h a l
o a l a h
a l a h
a h a l
a h o l
a l o h
o l a h
o h a l
In three letter words we cannot include all the consonants.
we cannot have any three letter with the condition that "The word should start with a vowel and no "TWO" consonants should be together"

Last edited: Dec 20, 2012
6. Dec 20, 2012

### pasmith

I would interpret "all the letters" to mean all of them: two A's, an 'H', an 'L' and an 'O'. But your interpretation (which I would phrase as "any of the letters") does reproduce the given answer.

7. Dec 20, 2012

### haruspex

There's also VCVVC for the other 6.

8. Dec 21, 2012

### Vineeth T

Oops! I forgot to include that case in my previous post.
so delete all the 4 letter words and include the following:
a l a o h
a h a o l
a l o a h
a h o a l
o l a a h
o h a a l
Also forgive me for giving the wrong answer.