Some Counting Principal Questions! Help!!!?

Question

Can you tell me if I am correct on these??? Thanks.
a. A sporting goods store stocks 10 styles of shoes. Each style comes in 8 sizes and 5 colors. How many shoes must the store stock in order tohave one pair of each size?
(10)8(5) = 400???
b. A business women has 8 skirts and 4 sweaters. How many different outfits can she make? 4*8 = 32?
c. A menu contains 5 main dishes, 3 side dishes, 2 salads, and 4 desserts. How many different meals can be made with the section? 5x3x2x4?? 240???
d. How many ways are there of drawing an ace from one deck of cards and a diamond from a second deck of cards.?? Need help on this one.

Thanks

Answer 1

Yeah you have the hang of it.

Let me do the last one.
There are 4 ways of choosing an ace, and 13 ways of choosing a diamond. So combined, that's 4*13 = 52

Answer 2

a) Assuming this means "one pair of each style in each size and each colour", then you have the right number of pairs - but you need to multiply by 2 to get the number of shoes.

b) Correct.

c) Correct logic, but 5×3×2×4 = 120.

d) There are 4 aces and 13 diamonds in any standard deck of cards. So there are 4 ways to draw the ace and 13 ways to draw the diamond, for a total of 4×13 = 52 ways.
(The reason they are specified to be in different packs is that the answer changes if they are in the same pack and you don't put the ace back before drawing the diamond.)

Answer 3

d) Probability of drawing ace = 4/52
Probability of drawing a diamond = 13/52
Probability of drawing an ace and a diamond = (4/52)(13/52) = 52/2704 = 1/52

Others look OK.

Answer 4

d. 4/52 for the ace 13/52 for the diamond. so to get both the probability is (4/52)*(13/52)=1/52

Answer 5

a. right
b. right
c. it is 5x3x2x4 but that is equal to 120
d. ace: 1/13 diamond: 1/4