Author Topic: perms and combs question (Read 3465 times) Tweet Share

mathsbrain · « **on:** July 04, 2017, 12:37:31 am »

Given 50 cards with the integers 1, 2, 3, ... 50 printed on them, how many ways are there to select 9 distinct cards, such that no two cards have consecutive numbers printed on them?
Detailed explanation appreciated! Thanks

RuiAce · « **Reply #1 on:** July 04, 2017, 12:48:04 am »

This does not appear to be a 4U question at all to me. Please provide either the source of this question or the final answer so we have a direction to head towards.

mathsbrain · « **Reply #2 on:** July 04, 2017, 12:00:58 pm »

yeah its a 4unit question in the sydney boys 2012 trials. apparently the answer is 42C9 and they say place the 9 consecutive cards in the 42 spaces(including both ends) of the rest of the cards....not sure how that makes sense

RuiAce · « **Reply #3 on:** July 04, 2017, 01:19:14 pm »

The analogy that they probably used hit me earlier this morning but I was still tired and couldn't really see how it'd work. Gave it more thought now though and it begins to barely puzzle up.
$\text{Denote any unwanted cards as }\bigcirc \text{ and wanted cards as }\times$
$\text{We would like to have 41 unwanted cards. This will give us}\\ \underbrace{\bigcirc \bigcirc \bigcirc \dots\bigcirc\bigcirc \bigcirc}_{\text{41 of them}}$
$\text{The cards that we want must involve non-consecutive numbers}\\ \text{It is the cards that we want that are restricted. Not the cards that we don't want.}$
$\text{To ensure that our cards we want are non-consecutive, we need to ensure that}\\ \text{between every two }\times\text{'s there always exists (at least) one }\bigcirc\text{ between them}$
$\text{What the so called 'spaces' are}\\ \text{can be more easily visualised like this:}\\ \_\bigcirc \_ \bigcirc \_ \bigcirc \_ \bigcirc \dots \bigcirc \_ \bigcirc$
$\text{Essentially, we have }42\text{ underscores }(\_)\\ \text{In each of those }\_\text{, we may place }\textbf{at most}\text{ one }\times\\ \text{This ensures that no two cards we want are consecutive numbers.}$
$\text{42 underscores, 9 crosses. Hence }\binom{42}9$
The idea is to not treat the numbers as just ordinary numbers, but instead as positions in a line. We simply make sure no two consecutive positions are being chosen to ensure that the same thing happens with the numbers.

Note also that once we place down the 9 crosses we will have 41 circles and 9 crosses. This counts as one feasible possibility. And then we do the same thing to get all the other possibilities.

mathsbrain · « **Reply #4 on:** July 04, 2017, 03:05:42 pm »

wow that is very smart!!! so thats a 4unit question?

seems unbelivably hard...

RuiAce · « **Reply #5 on:** July 04, 2017, 03:30:13 pm »

I doubt I'd have gotten that one out on the day. I only got better at counting after first year uni when I learnt these techniques

This sort of stuff helps get state ranks

pikachu975 · « **Reply #6 on:** July 07, 2017, 11:25:36 pm »

Quote from: RuiAce on July 04, 2017, 03:30:13 pm

I doubt I'd have gotten that one out on the day. I only got better at counting after first year uni when I learnt these techniques

This sort of stuff helps get state ranks

Do you have any resources to learn these techniques?

RuiAce · « **Reply #7 on:** July 08, 2017, 09:24:50 am »

Quote from: pikachu975 on July 07, 2017, 11:25:36 pm

Do you have any resources to learn these techniques?

I don't know if I'm allowed to distribute my MATH1081 materials on here (university copyright issues)

That trick isn't really a technique that can be taught. It comes from practicing ways of thinking about combinatorics from a different perspective.

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Author Topic: perms and combs question (Read 3465 times) Tweet Share

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