combinatorics questions on combinations and permutations

Question

1) A car dealership has seven vehicles each of a different color to line up on display in the showroom. Three of them are sedans and four of them are SUVs.

a. How many ways are there to line up the vehicles on display?

I'm not sure if the answer is $7!$ or if it is $\frac{7!}{3!4!}$ because there are 3 sedans and 4 SUVs

b. How many ways are there to line up the vehicles if all of the sedans must be together?

If al the sedans must be together than you have (SSS)VVVV Now I have five 'objects'. I'm confused if the answer is just $5!*3!$ or if it is $\frac{5!}{4!}*3!$ since the 4 comes from the 4 SUV's 'repeating' each other. and the $3!$ comes from the way the 3 sedans need to permute each other. The reason I'm confused is because I don't know if the four SUV's are 'repeating' each other because they are different colors.

c. If the black SUV and the red sedan must be next to each other, how many ways are there to line up the vehicles?

(BR)SSVVV

So is it $\frac{6!}{2!3!}*2!$

2) There are 12 points on a piece of paper. No three of them are on the same line.

a. How many different triangles can you draw using these points as vertices?

$12\choose 3$ Is this right?

b. How many different quadrilaterals can you draw using these points as vertices?

$12\choose 4$ or this ?

3) In how many ways can you write 18 as the sum of three counting numbers?

$18 P 3=4896$ or this?

Answer 1

1(a) Answer is 7! as each car is distinct and is of different color. 1(b) Answer is 5!3! same reason 1(c) Answer 6!2! 2(a),(b) are correct And for the third one consider non-negative integral solution of the equation x+y+z=18. Consider three distinct case I: x=y=z which is one solution 6+6+6 Case II: Two of them equal and third unequal say 5+5+8 and Case III: solve the equation when x

Answer 2

1a) It is $\frac{7!}{4!3!}$ if the 3 sedans is identic as well as 1the 4 SUVs. But, since it is different, then the answer is $7!$.

1b) Think of the 3 sedans as a group. Now, there are 5 objects (a group of sedans, three SUVs). So, there are $5!$ ways to arrange them. Since the 3 sedans can be arranged inside the group, we must multiply it by $3!$. Hence, the total number of ways are $5!3!$.

1c) Just like the previous part, think of the black SUV and red sedan as a group. The answer is $6!2!$.

Your answers for number 2 are correct.

As suggested in the comment, it is better to ask the number 3 separately.