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January 18, 2020, 01:15:49 am

### AuthorTopic: Permutations Questions Help!  (Read 374 times)

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#### Twisty314

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##### Permutations Questions Help!
« on: February 03, 2019, 12:16:23 pm »
+1
Hey everyone! Doing permutations at the moment and I found some questions very difficult. There are  answers to each question (which are in spoilers), so I really just need the solution. Anyone who can answer any of my questions: thanks so much!

Simiplify:$\frac{1}{n!}+\frac{1}{\left(n+1\right)!}$ Apparently it has got something to do with the identity: $n!=n\cdot \left(n-1\right)!$
Spoiler
$\frac{n+2}{\left(n+1\right)!}$

How many ways can 3 potatos be put in 5 boxes if every box can get more than 1 potato?
Spoiler
125

There are 4 differently coloured shirts. How many arrangements are possible, if you use at least 2 shirts arranged in a row?
Spoiler
60

You have 6 different colours to paint a cube with. If you are looking at on face androtate the cube and get the same colour, it is considered the same colour. How many ways can you paint this cube, so all of its faces have different colours?
Spoiler
30

How many ways can 3 onions and 2 zucchini be arranged in a row if they alternate?
Spoiler
12

The numbers 0 to 5 inclusive must be arranged without repeat. How many ways is this possible, if:
a) it is a 4 digit number, able to be divisible by 5?
b) it is a number < 6000?
c) it is a 3-digit number that is even?

Spoiler
108, 431, 52 respectively

How many palindromic numbers can you make with 5 and 6 digits?
Spoiler
900, 900 respectively

Thanks all!
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#### fun_jirachi

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##### Re: Permutations Questions Help!
« Reply #1 on: February 03, 2019, 10:49:48 pm »
+5
Hey there!

For the first question, you use that identity in conjunction with a bit of algebra to work your way around.

$\frac{1}{n!}+\frac{1}{(n+1)!}=\frac{(n+1)!+n!}{n!(n+1)!}=\frac{n!(n+2)}{n!(n+1)!}=\frac{n+2}{(n+1)!}$

Second question is reasonably simple. Since you can put each potato in any of the five boxes, and it doesn't matter how many each gets, you have 5 boxes to put in the first potato, 5 boxes to put in the second potato, 5 boxes to put in the third potato. So the answer is just 53 or 125.

Third question: You have three ways of arranging them: 2 rows of 2, 1 row of 3 and 1 of 1, 1 row of 4. With the 2 rows of two, effectively what you need to do is change the diagonals ie. where the Y's or X's are. You need to change what is diagonally adjacent to a fixed shirt, since basically most of the other combos are just rotations. for this part you get 3x2!x2! (try drawing them out, it'll make sense. theres only 12 anyway). Then, for the row of 3 and the row of 1 it's just 3! (for the row of 3) x 4 (for the different shirts in the row of 1) and the last is 4!. Add them all up and you should get 60.
YX
XY

Fourth question: not too sure, gonna come back to this one

Fifth question: Since you have 3 onions and 2 zucchini, they have to go in the order OZOZO. With this set order, it's easy to see that the answer will be 3!x2!, or 12.

Sixth Question:
a) Let the last number be 5. Then the first digit must be non-zero ie.1, 2, 3, 4. The second digit must be the three not used and zero, and the third must be the three not used. So with 5 as the last digit, there are 4x4x3x1 combos. With zero as the last digit, there are 5 possibilities for the first digit, four, then three for the respective digits. This becomes 5x4x3x1. Add those together, you get 48+60=108.
b) Basically everything with less than 5 digits
4 digits -- 5x5x4x3 --> the first number thing is for the non-zero, seen above in a)
3 digits -- 5x5x4
2 digits -- 5x5
1 digit -- 6 (we can include zero)
add them up, you should get 431.
c) Set the last digit to be a non zero even number (2 or 4). Then the first digit must be non-zero, and there are four options, then including zero with the other three leftovers, we get 4x4 x2(for the two different numbers). Then if we stick zero at the end, there is an additional option for the first digit, making it 5x4x1. Add them up and you should get 52.

Seventh question -- the intuition behind this is that the first number is the same as the last, second same as second last and so on. For the five digit number we have ABCBA. There are 9 options for A (A is a non-zero digit) while for B and C we have 10 options. 9x10x10=900. The same thing applies for the six digit number since you're effectively making it ABCCBA.

Hope this helps
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#### Twisty314

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##### Re: Permutations Questions Help!
« Reply #2 on: February 04, 2019, 06:28:29 pm »
+3
Woo thanks fun_jirachi!

I'm finding how to work out the algebra of the first question a bit tricky. Can you explain each step really thoroughly for me? Sorry I'm slow.

The second question makes so much sense now - it's sad to see I missed such a simple question! haha

Third question I am finding your explanation a bit confusing. Could I have some more guidance on this one too? Thanks!

Fifth question: your explanation helped a lot - thanks heaps!

Sixth question helps a lot! Great advice!

7th Q - that makes more sense now! Thanks!

So, really just need more guidance on the first and third question (the 4th question too when it comes through! haha - let me know how it goes!) Thanks for all your explanations; just made my life that bit easier.

Thanks!
Twisty314
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#### RuiAce

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##### Re: Permutations Questions Help!
« Reply #3 on: February 04, 2019, 06:41:40 pm »
+6
Q4 doesn't quite make sense looking at it actually.
"You have 6 different colours to paint a cube with. If you are looking at on face and rotate the cube and get the same colour, it is considered the same colour. How many ways can you paint this cube, so all of its faces have different colours?"
Are you sure there's no error in copy and paste? Seems to not tell me anything useful here.

Q1 does indeed use the identity you've stated. Note that similar to $n! = n(n-1)!$, we also have $(n+1)! = (n+1)n!$. Here's the same thing but with some annotations for convenience
\begin{align*} \frac{1}{n!} + \frac{1}{(n+1)!} &= \frac{(n+1)!}{n!(n+1)!} + \frac{n!}{n!(n+1)!} \tag{common denominator}\\ &= \frac{ (n+1)! + n!}{n! (n+1)!} \tag{combining the fractions}\\ &= \frac{(n+1) n! + n!}{n!(n+1)!}\tag{using the identity}\\ &= \frac{n! [(n+1)+1]}{n!(n+1)!} \tag{n! is a common factor on the top}\\ &= \frac{n+2}{(n+1)!}\tag{cancel and simplify}\end{align*}
I'm taking a review at his Q3 in the meantime Edit: I think I interpreted the question differently to how he did actually. What I interpreted was that you have 4 shirts to start with. And then, you only want to arrange 2 in a row, 3 in a row, or 4 in a row.

For this kind of question, we do something similar to Q6 in that we broke up the cases.
$\text{In the case of 2 shirts in a row, we can choose 2 out of 4 shirts in }\binom{4}{2}\text{ ways.}\\ \text{Once we have the shirts, there are }2!\text{ ways of arranging them.}\\ \text{So this case yields }\binom{4}{2}2! = 12\text{ arrangements.}$
$\text{In the case of 3 shirts in a row, we can choose 3 out of 4 shirts in }\binom43\text{ ways}.\\ \text{Since we now have three shirts, there are then }3!\text{ ways of arranging them.}\\ \text{So this case yields }\binom43 3! = 24\text{ arrangements.}$
$\text{In the case of 4 shirts in a row, we choose all 4 out of 4 shirts in }\binom44\text{ (i.e. one) way}.\\ \text{If we need to arrange all four shirts, we have }4!\text{ possible arrangements.}\\ \text{So this case yields }4! = 24\text{ arrangements.}$
Adding them up, $12+24+24 = 60$ as required.

(In the future, maybe post fewer questions in the one go. Can be a bit exhausting looking a pile, aha)
« Last Edit: February 04, 2019, 06:52:44 pm by RuiAce »

#### Twisty314

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##### Re: Permutations Questions Help!
« Reply #4 on: February 04, 2019, 07:20:49 pm »
+1
Woo, my first reply from the legendary RuiAce! haha

Quote
Q4 doesn't quite make sense looking at it actually.
"You have 6 different colours to paint a cube with. If you are looking at on face and rotate the cube and get the same colour, it is considered the same colour. How many ways can you paint this cube, so all of its faces have different colours?"
Are you sure there's no error in copy and paste? Seems to not tell me anything useful here.

Yeah I was trying to rephrase it from the textbook, but here's the exact question:
Given six different colours, how many ways can you paint a cube so that all the faces have different colours? Two colourings are considered to be the same when one can be obtained from the other by rotating the cube.

Hopefully that worked. Maybe I misinterpreted the question so maybe try to explain the question itself to me as well? haha

Q1: Yep - that definitely cleared it up for me. Thanks so much! I see where I was getting confused. Such an annoying error! haha

Q3: Yep that helps a lot, thanks!

Quote
(In the future, maybe post fewer questions in the one go. Can be a bit exhausting looking a pile, aha)
Sorry!  Will keep this in mind next time I need help. Thanks so much Rui!
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#### RuiAce

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##### Re: Permutations Questions Help!
« Reply #5 on: February 04, 2019, 08:23:23 pm »
+6
Ah ok I see the discrepancy. Just take careful note of the word 'colourings'. A colour is just, say, red. But a 'colouring' is just like an arrangement of people - it is actually a configuration of multiple colours in some manner. (Here, they're basically the colours of the cube.)

This problem is in my opinion substantially harder than the average Year 12 combinatorics problem. I had some close ideas but I decided to research it out instead. It essentially acts similar to arranging objects in a circle in that a cube doesn't really have a 'start' either. (Note that for the circle, we could just affix a start and do $(n-1)!$ from there, or we just arranged all $n$ objects but recognised the $n$ different rotations, which resulted in $\frac{n!}{n} = (n-1)!$.)

I'll state two methods here. Both of them build on similar ideas with the circle. (Suppose our colours are A, B, C, D, E and F.)

Approach 1
To create some kind of 'start' (or reference point), colour the bottom face A. There are now $5$ colours for the top face.

Let's suppose, for a particular example now, the top side was coloured B. We now have 4 faces left (the front and side faces) to colour with C, D, E or F. But due to rotational symmetry of those 4 faces, they essentially behave exactly like a circular arrangement of 4 objects. This can be done so in $3!$ ways.

This gives a final answer of $5\times 3! = 30$.

Remark: Here, I chose to arrange the top face before dealing with the front and side faces. I could've dealt with the front and side faces first instead: choose 4 out of 5 colours and arrange the 4 in a circle: $\binom{5}{4} 3!$. And then there's only one colour left for the top face: $1$. This would give a final answer of $\binom54 3! \times 1 = 30$.

Approach 2
Commence by simply allocating all six colours without fear of rotation. This can be done in $6!$ ways.

Now we need to divide out the number of rotations that create repeated-counting. First note that colour A can occupy $6$ different positions, each of which would create rotational symmetry.

After that, for a particular example, suppose that A is again now the bottom face. The remaining symmetry left in the cube turns out to be the same as the circular arrangement. That is, we can then rotate the cube $4$ times to still get the same arrangement and force A to stay on the bottom. This is the other factor that contributed to our double-counting problem.

So by this approach, we'd have a final answer of $\frac{6!}{6\times 4} = 30$.
« Last Edit: February 04, 2019, 08:30:26 pm by RuiAce »

#### Twisty314

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##### Re: Permutations Questions Help!
« Reply #6 on: February 04, 2019, 09:44:48 pm »
+3
Quote
Ah ok I see the discrepancy. Just take careful note of the word 'colourings'. A colour is just, say, red. But a 'colouring' is just like an arrangement of people - it is actually a configuration of multiple colours in some manner. (Here, they're basically the colours of the cube.)

This problem is in my opinion substantially harder than the average Year 12 combinatorics problem.

Oh, woops! I'll take note of that. Thanks for pointing that out for me!

Really?! That's funny because it is actually a question in my spesh units 1/2 textbook! (I'm doing year 11 atm.)

Wow - you've really made sense of this for me. Both approaches help so much and the circle analogy is fantastic! Thank you so much Rui! Really appreciate your time.
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#### RuiAce

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##### Re: Permutations Questions Help!
« Reply #7 on: February 04, 2019, 10:37:12 pm »
+4
(Oh haha oops, when I said Year 12 I really meant Year 11/12.)

But otherwise cool