Sum of largest and smallest elements in a subset

Question

In how many subsets of {1,2,3,4,n}\{ 1, 2, 3, 4, \ldots n\} is the sum of the largest element and the smallest element equal to (n+1)(n+1)?

Solution

I am able to deduce the following:

  • If the smallest element is 11, then the largest element has to be nn. The rest of the n2n-2 elements can be "in" or "out", thus the total will be 2n22^{n-2}
  • Similarly, If the smallest element is 22, then the largest element has to be n1n-1. And again, it will have 2n42^{n-4} such subsets
  • \ldots \ldots

Final Answer

  • If nn is odd the total subsets will be: 2n2+2n4++212^{n-2} + 2^{n-4} + \ldots + 2^{1}
  • If nn is even the total subsets will be: 2n2+2n4++202^{n-2} + 2^{n-4} + \ldots + 2^{0}
#Combinatorics

Note by Mahdi Raza
12 months ago

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1 vote

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Comments

Can anyone suggest a cleaner and short-hand form for the above answer?

Mahdi Raza - 12 months ago

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This is good. Nothing to improve on.

Pi Han Goh - 12 months ago

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Ok.

Mahdi Raza - 12 months ago

The best answer possible is given. You can also write it as: nk2:2k2nk\sum_{n\geq k \geq2:2|k}2^{n-k}

Zakir Husain - 12 months ago

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Hmm.. it is pretty compact but could be hard to understand. Thanks though!

Mahdi Raza - 12 months ago

Pretty good!

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Thanks!

Mahdi Raza - 12 months ago

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No problem!

I had a thought:

For any integer xx and any prime pp, can

xp\frac{x}{p}

always be irreducible if xx isn't a factor of pp?

Make a note on it if you have proof.

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@A Former Brilliant Member If you know the definition of a prime number, this should be obvious to prove/disprove.

Pi Han Goh - 12 months ago

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@Pi Han Goh Yes. And I think the question should say "x isn't a multiple of p" in which case the answer should obviously be a yes unless I am missing out something very trivial. Here is a simple deduction

Let xp=k\frac{x}{p} = k where 'kk' is an integer.

For kk to be an integer, x=pkx = pk

That means xx has to obviously be a multiple of pp.

Mahdi Raza - 12 months ago

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@Mahdi Raza Great! You did see my comment to @Hamza Anushath?

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