quick question

[fredrick]

i was having trouble with this question:
a student committee of 5 is to be chosen from 4 male and 6 female students. how many committees could be selected which contain more female than male students??

cheers

[Collin Li]

i was having trouble with this question:
a student committee of 5 is to be chosen from 4 male and 6 female students. how many committees could be selected which contain more female than male students??

cheers

Okay, so you can have:
5 females, 4 females or 3 females (any less would result in less females than males, and then the feminists would kill us  :shock: )

Combinations = (6 C 5)(4 C 0) + (6 C 4)(4 C 1) + (6 C 3)(4 C 2)

The first bracket in each term is talking about how we can choose women, the second bracket is talking about how we can choose men.

[fredrick]

thanks!

[fredrick]

1 more
in how many ways can a group of 6 eople be divided into:
a)two equal groups
b)two unequal groups

[Collin Li]

1 more
in how many ways can a group of 6 eople be divided into:
a)two equal groups
b)two unequal groups

a) two equal groups: groups of 3
= (6 C 3)(3 C 3) = 6 C 3
I used (3 C 3) because by the time you select the second group, there is only 3 people left to choose from.

b) two unequal groups: groups of 1 and 5 and groups of 2 and 4
= (6 C 5)(1 C 1) + (6 C 4)(2 C 2)
= (6 C 5) + (6 C 4)

Alternatively:
= (6 C 1)(5 C 5) + (6 C 2)(4 C 4)
= (6 C 1) + (6 C 2)

But that's because 6 C 1 = 6 C 5, and 6 C 4 = 6 C 3, they're sort of symmetrical.

[fredrick]

thanks

[fredrick]

one last question:
cos(x)-sin(x)=1/4, then cos(x)sin(x) equals?

[Collin Li]

one last question:
cos(x)-sin(x)=1/4, then cos(x)sin(x) equals?

Ouch, that one is a toughie. You have to square both sides (it's really hard to think of this, in my opinion):
[ cos(x) - sin(x) ]^2 = (1/4)^2
=> cos^2(x) - 2cos(x)sin(x) + sin^2(x) = 1/16
=> [ cos^2(x) + sin^2(x) ] - 2cos(x)sin(x) = 1/16
=> 1 - 2cos(x)sin(x) = 1/16
=> 2cos(x)sin(x) = 15/16
=> cos(x)sin(x) = 15/32