Waste less time on Facebook — follow Brilliant.
×

Combinatorics (Thailand Math POSN 1st elimination round 2014)

Write a full solution.

1.1) Find the number of ways that the word GIGGLING can be arranged such that the words can't start with vowels.

1.2) There're 50 numbers from 1 to 50. How many different ways can it be arranged in straight line such that odd numbers are arranged in increasing order and 2,4,6,8 must be next to each other (can be any order).

2.) (I feel like there's a problem in this question) Let \(L(n,r)\) be the number of ways of setting \(r\) people sit on \(n\) identical round tables such that each tables have at least \(k\) people. Prove that

\[L(n,r) = (r-1)L(n,r-1) + \displaystyle \frac{(r-1)!}{(r-k)!}L(n-1,r-k)\]

for all natural numbers \(r,n,k\) and \(nk \leq r\)

3.) Let \(n\) be natural numbers, prove these by combinatorial proof.

3.1) \(\dbinom{2(n+1)}{n+1} = \dbinom{2n}{n+1} + 2\dbinom{2n}{n} + \dbinom{2n}{n-1}\)

3.2) \(\displaystyle \frac{\left((n+1)!(n^{2}+5n+7)\right)!}{((n+1)!)^{n!}\left((n+3)!\right)!}\) is an integer.

4.) There're \(20\) different gifts, give them to \(4\) of the \(8\) students (not necessary every gifts are given). If each \(4\) students get more than \(1\) gifts, find the number of ways to do that.

5.) There're \(n\) people in a group (Brilli is one of them). Choose at least \(2\) people and have them stand on a straight line. The rest of people, except Brilli, are sitting on a circular table. Find the number of ways to do that.

Check out all my notes and stuffs for more problems!

Thailand Math POSN 2013

Thailand Math POSN 2014

Note by Samuraiwarm Tsunayoshi
2 years, 11 months ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

3.1) Consider \( 2n \) footballers, a cricketer "A" and a tennis player "B". We have to select \( n+1 \) of them.

DIRECT METHOD :- Total number of ways is \( {2n + 2 \choose n+1} \).

INDIRECT METHOD :- There are four cases, (a) Both A and B are selected (b)Only A is selected (c) Only B is selected (d) Neither A nor B is selected.

Number of ways in case (a) = \( {2n \choose n-1}\).Since A and B are chosen, we have to select the other \( n-1 \) players from the footballers.

Number of ways in case (b) = \( {2n \choose n}\)..Since only A is chosen, we have to select the other \( n \) players from the footballers.

Number of ways in case (c) = \( {2n \choose n} \).Since only B is chosen, we have to select the other \( n \) players from the footballers.

Number of ways in case (d) = \( {2n \choose n+1} \).Since neither A nor B is chosen, we have to select the other \( n+1 \) players from the footballers.

Since both the methods must give the same answer, we can conclude that \( {2n+2 \choose n+1} = {2n \choose n+1} + {2n \choose n} + {2n \choose n} + {2n \choose n-1} \)

Siddhartha Srivastava - 2 years, 11 months ago

Log in to reply

Nice! You can also use Pascal's identity easily.

Samuraiwarm Tsunayoshi - 2 years, 11 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...