1. ## 4 Questions

Question 1:
Find the number of ways in which five balls of different color can arrange in a row.

Question 2:
A teacher has to select 5 boys from section A, 4 boys from section B and 3 boys from section C for a competition, where section A has total 8 boys, section B has total 7 boys and section C has total 6 boys. How many choices does the teacher have?

Question 3:
A variable name in a programming language must start with an alphabet or characters “_”, next letters can be an alphabet or number. Maximum size for the variable name is 5 characters. How many different variable names are possible?

Question 4:
In how many ways can nine students be partitioned into three teams containing four, three, and two students respectively?

2. These are the simplest possible counting problems concerning queues.
What have you done on any of them? Where is the difficulty?
Tell us what you do not understand.

3. ## Thanks

I got this assignment, I know that the answers are very simple, but I wanted to confirm that these are really so simple that even the answer of the first question is of one line?
5 x 4 x 3 x 2 x 1 so 120 is it fine?

4. Right ! It's the permutation of 5 .

5. It's been a while since these questions were asked and I'm really bored, so I'll take a stab at one of the questions. Hopefully i don't embarrass myself too badly.
Originally Posted by aamirpk4
Question 2:
A teacher has to select 5 boys from section A, 4 boys from section B and 3 boys from section C for a competition, where section A has total 8 boys, section B has total 7 boys and section C has total 6 boys. How many choices does the teacher have?
The teacher has ${8 \choose 5} + {7 \choose 4} + {6 \choose 3}$ choices

6. Hello, aamirpk4!

4)In how many ways can nine students be partitioned into three teams
containing four, three, and two students, respectively?
Answer: . ${9\choose4,3,2} \:=\:\frac{9!}{4!\,3!\,2!} \:=\:1260$

7. Originally Posted by aamirpk4
Question 3:
A variable name in a programming language must start with an alphabet or characters “_”, next letters can be an alphabet or number. Maximum size for the variable name is 5 characters. How many different variable names are possible?
+the First case: the first letter is alphabet .
-The first letter has 26 choices
-The 2sd letter has 36-1 choices (26 alphabetic characters and 10 numbers)
-The 3rd letter has 36-2 choices.
-...
-...
So , we have :26 + 26*35 + 26*35*34 + 26*35*34*33 + 26*35*34*33*32 =: S1 choices for this case.

+the Second case: the first letter is "_" character.
-The first has 1 choice.
-The 2sd letter has 36 choices.
-The 3rd letter has 36-1 choices.
-...
So , we have :36 + 36*35 + 36*35*34 + 36*35*34*33 =: S2
(considering that "_" is not a variable name)

The final result is S1+S2 .

8. Originally Posted by Jhevon

The teacher has ${8 \choose 5} + {7 \choose 4} + {6 \choose 3}$ choices
I might be wrong, but should the combinations be multiplied instead of added?

9. Originally Posted by DivideBy0
I might be wrong, but should the combinations be multiplied instead of added?
Not when you divide the problem into sub-cases.

For example, say the problem goes something like this "...find how many combinations for at least 3 students..."

You can simplify the problem by finding:
exactly 1 student
exactly 2 students
exactly 3 students

And then add up the combinations in all the sub-cases.

10. Originally Posted by DivideBy0
I might be wrong, but should the combinations be multiplied instead of added?
Question 2:
A teacher has to select 5 boys from section A, 4 boys from section B and 3 boys from section C for a competition, where section A has total 8 boys, section B has total 7 boys and section C has total 6 boys. How many choices does the teacher have?

Yes you are correct we must multiply and not add to answer this partiucular question.

11. Originally Posted by Soroban
Hello, aamirpk4!

Answer: . ${9\choose4,3,2} \:=\:\frac{9!}{4!\,3!\,2!} \:=\:1260$

Hi Soronban.

Another way :
We choose 4 students from 9 students : ${9\choose4}$
After that choosing 3 students from 5 students : $5\choose3$
And the final , choosing 2 from 2 : $2\choose2$
So , the result is : ${9\choose4}$* $5\choose3$* $2\choose2$ = 1260.

@ Soronban : Would you like to tell how we receive the function that you gave ? I mean that what way you think about the question 4?

12. Originally Posted by DivideBy0
I might be wrong, but should the combinations be multiplied instead of added?
Originally Posted by Plato
Question 2:
A teacher has to select 5 boys from section A, 4 boys from section B and 3 boys from section C for a competition, where section A has total 8 boys, section B has total 7 boys and section C has total 6 boys. How many choices does the teacher have?

Yes you are correct we must multiply and not add to answer this partiucular question.
Indeed!

looking back at this, i have no idea why i decided to add...