Combinatorics (1st math Thailand POSN 2014)

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

  • 1.1) Find the coefficients of x5x^{5} from the expansion of (2+x+x2)8(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) k=0r(rk)2=(2rr)\displaystyle \sum\limits_{k=0}^{r} \dbinom{r}{k}^{2} = \dbinom{2r}{r} for any natural number rr.

  • 2.2) (9n)!25n3n\displaystyle \frac{(9n)!}{2^{5n}3^{n}} is always integer for any natural number nn.

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)V(r,n) be the number of ways of putting rr different objects into nn identical boxes such that each boxes must have at least kk objects Prove that

V(r,n)=nV(r1,n)+(r1rk)V(rk,n1)V(r,n) = nV(r-1,n) + \dbinom{r-1}{r-k}V(r-k,n-1)

for any natural number r,n,kr,n,k and rnkr \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
5 years, 1 month ago

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3) You are looking for the number of solutions to

a+b+c+d+e+f=15 a + b + c + d + e + f = 15 , subject to 1a3 1 \leq a \leq 3 , and similarly for all the other variables.

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

Calvin Lin Staff - 5 years, 1 month ago

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Wow, this is nice! Thank you ^_^

Samuraiwarm Tsunayoshi - 5 years, 1 month ago

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

Samuraiwarm Tsunayoshi - 5 years, 1 month ago

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M={4na1,2na2,2na3,1na4}M =\{ 4n\cdot a_{1}, 2n\cdot a_{2}, 2n\cdot a_{3}, 1n\cdot a_{4}\} such that M=9n|M| = 9n.

Number of permutations = (9n)!(4!)n(2!)n(2!)n(1!)n=(9n)!25n3n\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.

Samuraiwarm Tsunayoshi - 5 years, 1 month ago

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