Question

In a book fair, Joyce wants to buy 6 novels, 3 comics, 2 traveler's guides and 3 cookbooks. However, she can only buy 4 of them due to her budget. How many choices does she have if:
(a) she buys 3 novels and 1 book of another type?
my answer: 6C3*8C1=160

(b) she buys 2 novels and 2 books of different other types?
my answer : 6C2*8C2=420

@janaki.t23 @wio @douglaswinslowcooper @sourwing

Answer 1

a is right. can you explain how you got your answer for b ?

Answer 2

still here ?

Answer 3

@janaki.t23 as for part b, it said she buys 2 novels, then 6C2. And 2 books of different other types. 3 comics+2 traveler's guides+3 cookbooks=8. From these 8 books, get 2. then 8C2...
this is my way to do it..

Answer 4

upto 6C2 its correct. remaining is wrong

Answer 5

3C2*2C2*3C2?

Answer 6

yes something like that, but still that expression makes no sense.
she wants to buy 2 books of DIFFERENT other types

Answer 7

ohh, that means for example:she wants to buy 1 cookbook and another book should not be cookbook. right?

Answer 8

thats right

Answer 9

total ways to pick 2 books from available remaining 8 books = 8C2
total ways to pick 2 books of SAME type(both comics or both traveller guides or both cookbooks) = 3C2 + 2C2 + 3C2

total ways to pick 2 books of DIFFERENT types = 8C2 - [3C2 + 2C2 + 3C2]

Answer 10

multiply this with novels

Answer 11

ohh, i almost forget we can use subtraction

Answer 12

total ways to buy 2 novels and 2 books of different other types = 6C2[8C2 - (3C2 + 2C2 + 3C2)]

Answer 13

"Different other types" seems to have been the joker.