Combinations (1 Viewer)

[zeek]

A committee of 6 is to be selected from 10 people of whome A and B are two. How many committees can be formed if A is excluded but B is included?

My working goes something like this:

A is excluded therefore there are 9 people to choose.
B is included therefore there are 5 places to fill.
.: ⁹C₅ = 126
but the answer says 140....

 

[webby234]

Well I would have thought 8C5, because you are choosing 5 from the remaining 8 (having already removed A and included B). But that's certainly not 140 so i'm not sure where we've gone wrong.

The only way i can see you getting 140 is
8C4 * 2 = 140

but i'm not sure why that would be the correct answer

 

[zeek]

alrite well here's another one:

In how many ways can n things be shared between 2 people?

The answer is 2ⁿ - 2

EDIT: nvm i got the answer

 

[zeek]

and some more lol...

The ratio of the number of combinations of (2n+2) different objects taken n at a time to the number of combinations of (2n-2) different objects taken n at a time is 99:7. Find the value of n.

Five cards are drawn at random from a pack of 52 playing cards. What is the probability that they are all from the same suit? Answer:33/16660

Five cards are drawn from a pack of 52 playing cards. What is the probability of drawing at least 3 aces? Answer: 19/10829

 

[Riviet]

The ratio of the number of combinations of (2n+2) different objects taken n at a time to the number of combinations of (2n-2) different objects taken n at a time is 99:7. Find the value of n.

{(2n+2)Cn}/{(2n-2)Cn} = 99/7

Solve for n from here.

Five cards are drawn at random from a pack of 52 playing cards. What is the probability that they are all from the same suit? Answer:33/16660

Probability = (52/52) x (12/51) x (11/50) x (10/49) x (9/48)

Five cards are drawn from a pack of 52 playing cards. What is the probability of drawing at least 3 aces? Answer: 19/10829

Probability = p(3 aces) + p(4 aces)

 

[webby234]

Yeah - for part 1 (good practice )

(2n+2)Cn/(2n-2)Cn
= [(2n + 2)!/n!(n + 2)!]/[(2n - 2)!/n!(n - 2)!]
= [(2n + 2)!(n - 2)!]/[(2n - 2)!(n + 2)!]
= [(2n + 2)(2n + 1)2n(2n - 1)]/(n - 1)n(n + 1)(n + 2)
4(2n + 1)(2n - 1)/(n - 1)(n + 2) = 99/7
28(4n² - 1) = 99(n² + n - 2)
112n² - 28 = 99n² + 99n - 198
13n² - 99n + 170 = 0
n = 99 +/- rt(99² - 8840)/26
n = (99 +/- 31)/26
n = 5, 68/26

Must be whole number so n = 5

Is there a simpler way? Way too much algebra.

 

[SoulSearcher]

I haven't seen a simpler way.

 

[zeek]

Urn A contains 6 white and 4 black balls. Urn B contains 2 white and 2 black balls. From urn A two balls are selected at random and placed in urn B. From urn B two balls are then selected at random. What is the probability that exactly one of these two balls is white? Answer: 128/225

 

[Riviet]

Urn A contains 6 white and 4 black balls. Urn B contains 2 white and 2 black balls. From urn A two balls are selected at random and placed in urn B. From urn B two balls are then selected at random. What is the probability that exactly one of these two balls is white? Answer: 128/225

Probability = (no. of desirables)/(no. of possible)

={no.(2W & 0B) + no.(1W & 1B) + no.(0W & 2B)}/(10C2.6C2)

=(6C2.4C1.2C1 + 6C1.4C1.3C1.3C1 + 4C2.2C1.4C1)/675

=128/225 #

Note: no.(event) means number of ways of choosing those coloured balls from urn A and putting them into urn B and drawing 1W and 1B from urn B (in each case).