I need help with these probability problems.

Question

1). Five chips are selected from a bag without replacement. The bag originally contained 6 yellow chips and 8 red chips. In how many ways can you choose 5 chips from the bag?

I did 14 C 5 and got 2002, is that correct.

2). In how many ways can you choose no yellow chips?

Answer 1

no

you have to look at the results,
you have 5 chips selected, that could be one of the following cases :
0Y 5R .... RRRRR, only 1 way OR 5!/0!5!)
1Y 4R .... YRRRR, OR RYRRR, ... 5!/4! = 5
2Y 3R .... 5!/(2!3!) = 10
3Y 2R .... 5!/(3!2!) = 10
4Y 1R .... 5!/(4!1!) = 5
5Y OR .... 5!/(5!0!) = 1

total number of ways is 32

This is not a probability question, but rather based on the little formula
for the number of ways that you can arrange p things, q alike of one kind, and r alike of another kind, which is
p!/(q!r!)

2) the number of ways you can choose no yellow chips is 1, namely RRRRR

Had you asked "what is the probability of choosing no yellow chip in choosing any 5 chips from the above that would be

1/( 14C5 ) = 1/2002

Answer 2

Correction:
I said at the end
"Had you asked "what is the probability of choosing no yellow chip in choosing any 5 chips from the above that would be

1/( 14C5 ) = 1/2002 "

that should have been 1/32