# Combinatorics (1st math Thailand POSN 2014)

1.) Answer with factorial notations and simplify into numbers.

• 1.1) Find the coefficients of $x^{5}$ from the expansion of $(2+x+x^{2})^{8}$
• 1.2) There're 10 people, including A,B,C,D. If we set them arrange in the long bench, find the number of different ways are there, such that only 2 of A,B,C,D are sitting together.

2.) Prove these statements by combinatorial proof.

• 2.1) $\displaystyle \sum\limits_{k=0}^{r} \dbinom{r}{k}^{2} = \dbinom{2r}{r}$ for any natural number $r$.

• 2.2) $\displaystyle \frac{(9n)!}{2^{5n}3^{n}}$ is always integer for any natural number $n$.

3.) Throw 15 6-sided regular dice, find the number of different ways such that every 6 different sides are shown and no more than 3 same sides are shown.

4.) Let $V(r,n)$ be the number of ways of putting $r$ different objects into $n$ identical boxes such that each boxes must have at least $k$ objects Prove that

$V(r,n) = nV(r-1,n) + \dbinom{r-1}{r-k}V(r-k,n-1)$

for any natural number $r,n,k$ and $r \geq nk$.

5.) There was a rumour inside the group of 10 people. This rumour is spread by e-mail and continuously spread by following rules.

• First, there was only 1 people know about the rumour called rumour-er.
• Each e-mail can be either forwarded directly (exactly 1 people and can forward again) or people who receive copies (can be 0 people or any number of people, but cannot forward again)
• People who received the e-mail (by both ways) can know who sent the mail, and those are considered to be rumoured
• People who received by forwarding can only forward the mail once and only forward to people who are not rumoured.
• Rumour-er can send as many e-mails as he/she want.

How many ways are there if the e-mails are sent exactly 2 times, and how many ways if sent exactly 3 times?

This is the part of Thailand 1st round math POSN problems. Note by Samuraiwarm Tsunayoshi
6 years ago

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3) You are looking for the number of solutions to

$a + b + c + d + e + f = 15$, subject to $1 \leq a \leq 3$, and similarly for all the other variables.

Hint: Use the substitution $z = 3 - a$.

Staff - 6 years ago

Wow, this is nice! Thank you ^_^

2.) I use multiset to prove, but I forgot how to form a multiset.

$M =\{ 4n\cdot a_{1}, 2n\cdot a_{2}, 2n\cdot a_{3}, 1n\cdot a_{4}\}$ such that $|M| = 9n$.

Number of permutations = $\displaystyle \frac{(9n)!}{(4!)^{n}(2!)^{n}(2!)^{n}(1!)^{n}} = \frac{(9n)!}{2^{5n}3^{n}}$ which is always integer. Done!

I swear to goat that I haven't done this for years since the last year test.