Sum of largest and smallest elements in a subset

Question

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

Solution

I am able to deduce the following:

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

Final Answer

If $n$ is odd the total subsets will be: $2^{n-2} + 2^{n-4} + \ldots + 2^{1}$

If $n$ is even the total subsets will be: $2^{n-2} + 2^{n-4} + \ldots + 2^{0}$

#Combinatorics

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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

Mahdi Raza - 12 months ago

This is good. Nothing to improve on.

Pi Han Goh - 12 months ago

Ok.

Mahdi Raza - 12 months ago

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

Zakir Husain - 12 months ago

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

Mahdi Raza - 12 months ago

Pretty good!

A Former Brilliant Member - 12 months ago

Thanks!

Mahdi Raza - 12 months ago

No problem!

I had a thought:

For any integer $x$ and any prime $p$ , can

$\frac{x}{p}$

always be irreducible if $x$ isn't a factor of $p$ ?

Make a note on it if you have proof.

A Former Brilliant Member - 12 months ago

@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

@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 $\frac{x}{p} = k$ where ' $k$ ' is an integer.

For $k$ to be an integer, $x = pk$

That means $x$ has to obviously be a multiple of $p$ .

Mahdi Raza - 12 months ago

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

A Former Brilliant Member - 12 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$