Combinatorics Marathon!!

[Labib]

I got this Idea from AOPS!! But as Mr Zeits says(I'm not so sure about the spelling), shamelessly make others' ideas your own, so I did.

The rules are simple: (must read for people who dont know them)

1) I start with giving a problem
2) someone else posts the full correct solution of the problem and hides it. Then he posts another problem.
3) If his solution is somewhat wrong, another person posts the correct solution and the solution for the second problem and hides it. Then he posts another problem.
4) No one should post reply without full solution.

P.S.: this is the divisional olympiad period. so I hope the problems you post is easy and suitable for div olympiad.

so here goes!!

$P1$
Chelsea, Manutd and Arsenal have 33 players in their squads in total(each team has 11 players). How many ways can you form a team so that there are at least 1 player from each team in your team??

[Md. Alfaz Hossain]

3 players can be chosen from 3 team in (11C1)*(11C1)*(11C1)ways. Another 8 players can be chosen from remaining 30 players in 30C8 ways. So,the required team can be formed in (11C1)*(11C1)*(11C1)*(30C8) ways.

[Moon]

Alfaz: I guess the rule is to post a new problem with a solution.
BTW welcome to the forum. I think that it would be nice if you write mathematical expressions using LaTeX (see below)
Anyway new problem:
Problem 2:
A medical student has to work in a hospital for five days in January. However, he is not allowed to work two consecutive days in the hospital. In how many different ways can he choose the five days he will work in the hospital?

ref: wtc 3.19

[tushar7]

solution#
$15C_5+16C_5$ ..
i am not sure if my answer is correct or not .so not posting new problems or giving the full solution

[Moon]

Please, please, and please give complete solutions. Otherwise none of us is going to learn anything. I can't tell you whether your ans is correct or not because you haven't showed solution. Even if your ans is correct your solution might not be. So please send a complete solution, always!

[anumoshsad]

This is my first post to this forum
Let the chosen dates be $ x_{1}<x_{2}<x_{3}<x_{4}<x_{5}$. As there is no consecutive dates, after using the bijection $\{ x_{1},x_{2},x_{3},x_{4},x_{5}\}\leftrightarrow\{ x_{1},x_{2}-1,x_{3}-2,x_{4}-3,x_{5}-4\}$, we find the answer is $\dbinom{31-4}{5}=\dbinom{27}{5}$.
To understand the bijection clearly just think about the the example $\{1,3,5,7,9\}\leftrightarrow\{ 1,2,3,4,5\}$.

Anyway, its really cool to post to this forum. Good job, moon

[Zzzz]

Cool problem, cool solution

Here is another solution :
The student won't work on (31-5)=26 days. Now just place 5 plus signs (+) in these 27 places
\[\_1\_1\_1\_ ... \_1\_\]Here are 26 $1s$ and 27 spaces...

So the answer is $^{27}C_5$

For example : \[+1\_1+1+1\_1\_1+1\_1\_1+1\_1\_...\] means he will work on 1st January then he won't work for 2 days, then he won't work for 1 day, then for 3 days, then for 3 days.. then its over

Problem 3
How many 7-letter sequences are there which use only A,B,C (not necessarily all of those) with no A immediately followed by B, no B immediately followed by C, no C immediately followed by A ?

[Avik Roy]

Here is my solution:

Obviously, the first of the letters have three options to choose from. For each of the successive letters, we have two letters availabe. So the answer is $3.2^6$

P4
How many four digit numbers $\overline{abcd}$ are there so that $\overline{abcd} + \overline{bcda}$ is a four digit palindrome?

[A palindrome is a number that is the same if written in reverse order, like $1234321$]

[Labib]

Here's my solution for $P4$::

From the problem it's clear that the palindrome's first and last digit would be $(a+d)$ and $(a+b)$. now, there are two possibilities!!!
$(a+d)=(a+b)$ or $(a+d)=(a+b+1)$

It can easily be seen that if $(a+d)\geq(a+b)\geq10$ the number $abcd+bcda$ wouldn't be a four digit palindrome!!

Now, let's work with $(a+d)=(a+b)$. we can make a table about how many ways we can choose $b,c,a$.

we'll see that if we choose $b=n: 1\leq n\leq 8$

we'll have $9-n$ ways to choose $a$ and $10-n$ ways to choose $c$. thus we'll have $240$ possible palindromes.

now let's work with $(a+d)=(a+b+1)$

as we know,
if $(a+d)\geq10$, $abcd+bcda$ will not be a 4 digit palindrome.

so $a+d<10$. but at the same time $c+d\geq10$

if we make a table we'll see, if we choose $d=n:2\leq n\leq8$ we'll have $n$ ways to choose $c$ and $9-n$ ways to choose $a$.

so we'll have $112$ ways.

so in a total we have $240+112=352$ ways to form a 4 digit palindrome!!

please confirm if I'm right....

here's $P5$:::

From the set $\left \{ 1,2,....,10 \right \}$ how many ways can we choose 5 numbers so that their sum is $20$?

[Moon]

This is my first post to this forum
Let the chosen dates be $ x_{1}<x_{2}<x_{3}<x_{4}<x_{5}$. As there is no consecutive dates, after using the bijection $\{ x_{1},x_{2},x_{3},x_{4},x_{5}\}\leftrightarrow\{ x_{1},x_{2}-1,x_{3}-2,x_{4}-3,x_{5}-4\}$, we find the answer is $\dbinom{31-4}{5}=\dbinom{27}{5}$.
To understand the bijection clearly just think about the the example $\{1,3,5,7,9\}\leftrightarrow\{ 1,2,3,4,5\}$.

Anyway, its really cool to post to this forum. Good job, moon

Thanks for both your solution and your complement!
It is really nice to see you here, finally.