A newly-wed couple is using a website to design an eBook Wedding Album

total ways to choose 3 out of 6=6c3=20..
total ways to choose 19 out of 21=21c19=21*20/2=210..
so total ways=20*210=4200 ans B...

MAGOOSH OFFICIAL SOLUTION:

For the large photos, we need 6C3, which we calculated in the article:
6C3 = 20

For the smaller photos, we need 21C19, which by symmetry must equal 21C2, and we have a formula for that. In fact, in the article above, we already calculated that 21C2 = 210.

Now, by the FCP, we just multiply these: total number of possible albums = 20*210 = 4200.

Answer = B

Hi All,

We're told that a newly-wed couple is using a website to design an eBook Wedding Album to distribute to their friends and families. The template they have chosen has places for 3 large photos and 19 smaller photos. The couple has 6 large photos they could use for those three slots, and 21 smaller photos they could use for those 19 slots. We're asked for the number of different possible albums could they create. While the prompt does not explicitly state it, we are meant to assume that placing the same photos in a different order does NOT count as a "new" album, meaning that this is a Combination Formula question.

With the 6 large photos and 3 slots, there are 6c3 = 6!/3!3! = (6)(5)(4)/(3)(2)(1) = 20 possible groups of large photos

With the 21 small photos and 19 slots, there are 21c19 = 21!/19!2! = (21)(20)/(2)(1) = 210 possible groups of small photos

(20)(210) = 4200 possible albums

Final Answer:

GMAT assassins aren't born, they're made,
Rich

The number of different albums they could create is:

6C3 x 21C19 = 6C3 x 21C2 = [(6 x 5 x 4)/(3 x 2)] x [(21 x 20)/2] = 20 x 210 = 4,200

Answer: B