1. ## Problems

Help please, I suck at permutations and combination.

1.) A. In how many different ways can a student answer a 20 question true-false examination?
B. In how many ways can he answer the examination if he is to answer 18 out of 20 questions?

2. In how many ways can 4 male students, 4 female students and a professor sit in a row for class picture if:
A. No restrictions?
B. the professor must be in the middle
C. a certain student and the professor must occupy the ends?

3. Find the number of ways of selecting 6 balls from 6 red balls, 3 green ball and 5 black balls if each selection consists of 2 red balls, 2 green balls and 2 black balls.

4. In how many ways can 13 people be grouped into two groups of 8 and 5 if a certain person is to be in the larger group?

5. Ten people is to sit a round table. Three particular people insist on sitting next to each other. In how many ways can ten be seated?

Help please, hope you can help me. Thank you very much.

2. 1.
$\displaystyle A) 2^{20}$

$\displaystyle B) {20\choose 18} \times 2^{18}$

3. 2.
$\displaystyle A) 9!$
$\displaystyle B) 1!\times 8!$
$\displaystyle C) 2!\times 7!$

4. 3.
$\displaystyle {6\choose 2}\times{3\choose 2}\times{5\choose 2}$

4.
$\displaystyle {13\choose 1}\times{12\choose 7}\times {5\choose 5}$

5.
$\displaystyle 3!\times 7!^2$

5. Originally Posted by Anonymous1
3.
$\displaystyle {6\choose 2}\times{3\choose 2}\times{5\choose 2}$

4.
$\displaystyle {1\choose 1}\times{12\choose 7}\times {5\choose 5}$

5.
$\displaystyle 3!\times 7!^2$

what do u mean by the two numbers inside the parenthesis? divide them?

6. nCr. You know combinations..

$\displaystyle {n\choose k}= \frac{n!}{(n-k)!k!}$

7. Sorry to bother but can briefly explain some of the answers?

8. Which ones?

9. Originally Posted by Anonymous1
3.
$\displaystyle {6\choose 2}\times{3\choose 2}\times{5\choose 2}$

4.
$\displaystyle {13\choose 1}\times{12\choose 7}\times {5\choose 5}$

5.
$\displaystyle 3!\times 7!^2$
How did you get the (6 over 2) ?

10. We have six red balls. We want two of them. How many combinations of two red balls can we select when we have a group of six in total?

$\displaystyle 6$ choose $\displaystyle 2,$ or $\displaystyle {6\choose 2} = \frac{6!}{(6-2)!2!}$

11. The $\displaystyle {a \choose b}$ notation is just a notation for this purpose, it does not mean division. It means how many combinations of $\displaystyle b$ elements in $\displaystyle a$ elements exist.