Solving Permutation Questions: 4-Digit Even Numbers and Letter Arrangements

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The discussion focuses on solving permutation problems involving 4-digit even numbers and letter arrangements. For the first question, the total number of 4-digit even numbers greater than 5000 that can be formed using the digits 0, 1, 2, 3, 5, 6, 8, and 9 without repetition is calculated to be 420, with 120 of those ending in 0. The second question confirms that n_P_3 equals 120, leading to the conclusion that n is 6. The final question involves arranging the letters of "ALOHA" with specific conditions, resulting in 18 valid arrangements when vowels must start and consonants cannot be together. Overall, the thread provides detailed solutions and clarifications for each of the permutation problems presented.
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Homework Statement



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?

Homework Equations





The Attempt at a Solution


1. I am doing the cases, and i set it up as 1*7*6*2 + 3*7*6*3 which isn't right
answer is 420, b) is 120

2. n!/(n-3)! = 120 = n(n-1)(n-2)=120 doesn't make sense from here.
answer is 6

3. I was thinking 3*4*2*1 = 24 which isn't right either. answer is 18
 
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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
Final answer =240 + 180=420!

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!
 
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!
 
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.
 
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:
Vineeth T said:
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.

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.
 
pasmith said:
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.
There's also VCVVC for the other 6.
 
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.
 
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