Perms and combs (final set...) (1 Viewer)

[deswa1]

Hey guys,

This is the final set of questions that I couldn't do. Thanks so much:

a) The six faces of a number of identical cubes are painted in six distinct colours. How many different cubes can be formed?

b) A poker hand of five cards is dealt from a standard pack of 52. Find the probability of obtaining:
i) One pair
ii) Two pairs etc. - it goes on up to full house but these two should be fine and I should hopefully be able to get the rest from the working from these

c) Five diners choose randomly from a menu featuring five main courses. Find the probability that exactly one of the main courses is not chosen by any of the diners.

d) A group of n men and n+1 women sit around a circular table. How many arrangements are possible if no two men are to set next to one another?

Thanks

 

[barbernator]

a) using numbers 1-6. Set 1 as an arbitrary face. The face opposite can have 2-6. The rest of the faces are in a circular pattern so have 3! ways of arrangement. so 5 x 3! = 30 different dices.

b)i) 52C5 ways of choosing 5 cards. ways of choosing a pair. We can choose any kind, from 1-13, and must pick 2 of them. so 13 x (4C2). Now we must pick any other cards that are not of the same kind. so of the kinds, we have 12C3. These are the 3 different kinds we can pick, but within each kind, there are four cards.

multiplying these probabilities together, we have 13x(4C2) x 12C3 x 4^3.

dividing by number of ways, 52C5.

we obtain 352/833 = 42%~

ii) Similarly, we must choose 2 pairs. 13C2 x 4C2 x 4C2. Now we must pick 1 different card. x44.\

so 13C2 x 4C2 x 4C2 44 / 52C5 = 198/4165 = 4.7%~

c) Using binomial probability, their choices are (4/5 + 1/5)^5. The chance that none choose the meal is (5C0)(4/5)^5 = 32%~

d) In this circle, between each man there will be 1 woman, except between 1 pair, there will be 2. So by grouping 2 women together, we will have equal numbers of men and women and we can resolve the extra numbers at the end. So the arrangement of men is (n-1)! and the arrangement of women is (n-1)! but as there is one group of women, the different amount of perms of women able to be chosen are (n+1)P2. So the answer is ((n-1)!)^2 x (n+1)P2

Could you tell me if these are all correct

 

[deswa1]

a) using numbers 1-6. Set 1 as an arbitrary face. The face opposite can have 2-6. The rest of the faces are in a circular pattern so have 3! ways of arrangement. so 5 x 3! = 30 different dices.

b)i) 52C5 ways of choosing 5 cards. ways of choosing a pair. We can choose any kind, from 1-13, and must pick 2 of them. so 13 x (4C2). Now we must pick any other cards that are not of the same kind. so of the kinds, we have 12C3. These are the 3 different kinds we can pick, but within each kind, there are four cards.

multiplying these probabilities together, we have 13x(4C2) x 12C3 x 4^3.

dividing by number of ways, 52C5.

we obtain 352/833 = 42%~

ii) Similarly, we must choose 2 pairs. 13C2 x 4C2 x 4C2. Now we must pick 1 different card. x44.\

so 13C2 x 4C2 x 4C2 44 / 52C5 = 198/4165 = 4.7%~

c) Using binomial probability, their choices are (4/5 + 1/5)^5. The chance that none choose the meal is (5C0)(4/5)^5 = 32%~

d) In this circle, between each man there will be 1 woman, except between 1 pair, there will be 2. So by grouping 2 women together, we will have equal numbers of men and women and we can resolve the extra numbers at the end. So the arrangement of men is (n-1)! and the arrangement of women is (n-1)! but as there is one group of women, the different amount of perms of women able to be chosen are (n+1)P2. So the answer is ((n-1)!)^2 x (n+1)P2

Could you tell me if these are all correct

They're all correct except for d where the answers give: <a href="http://www.codecogs.com/eqnedit.php?latex=\frac{n!(n@plus;1)!}{(2n)!}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{n!(n+1)!}{(2n)!}" title="\frac{n!(n+1)!}{(2n)!}" /></a> and c where the answers have 48/125 or 38.4%

But thanks heaps bro- it's making sense now haha

 

[barbernator]

They're all correct except for d where the answers give: <a href="http://www.codecogs.com/eqnedit.php?latex=\frac{n!(n@plus;1)!}{(2n)!}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{n!(n+1)!}{(2n)!}" title="\frac{n!(n+1)!}{(2n)!}" /></a> and c where the answers have 48/125 or 38.4%

But thanks heaps bro- it's making sense now haha

oh whoops, for c, I neglected the fact that the other 4 must each be chosen at least once. but im not sure about that answer either, because if my method showed that at least 1 dish is definitely not chosen, then the percentage should be <32% hmm

and the answer for d from the textbook is a fraction lol, even though it says how many ways.