Another perm and comb question (1 Viewer)

[cutemouse]

Consider a pack of 40 cards consisting of the colours red, blue, yellow and green, each with cards numbered 1 to 10. A hand of five cards is dealt from the pack. Find the probabilities that the cards were in ascending order.

Thanks

 

[cutemouse]

Hi, another question:

Five couples are to be seated around a dinner table. What is the probability that each member of a couple ends up opposite their partner?

 

[scardizzle]

Consider a pack of 40 cards consisting of the colours red, blue, yellow and green, each with cards numbered 1 to 10. A hand of five cards is dealt from the pack. Find the probabilities that the cards were in ascending order.

Thanks

yeh i asked bakes about this question

the question is unclear what it means is having the cards in a straight
e.g. 12345,23456 etc.

therefore solution is 4C1x4C1x4C1x4C1x4C1 x no. of straights you can make

just write it out it, doesnt take too long

then divide this by 40C5

 

[ninetypercent]

Five couples are to be seated around a dinner table. What is the probability that each member of a couple ends up opposite their partner?

no of ways without restriction: 9! = 362880
no of ways with restriction:

Fix one person. The partner has one place to sit
the next person has 8 ways to sit.
the next has 6
the next has 4
the next has 2
the next has 1
= 8 x 6 x 4 x 2 x 1
= 384

Probability = 384/362880
= 1/945

 

[cutemouse]

the next person has 8 ways to sit.
the next has 6
the next has 4
the next has 2
the next has 1

Umm, why does it reduce by 2 each time?

Thanks

 

[ninetypercent]

Umm, why does it reduce by 2 each time?

Thanks

You have ten people you need to arrange.
A1, A2, B3, B4, C5, C6, D7, D8, E9, E10

Fix A1 in one spot. A2 MUSt sit opposite to A1, so A2 has only one spot to sit
B3 will have 8 spots left to sit, as A1 and A2 have already sat around the table
B4 will have one spot to sit, as he/she must always be opposite to B3.
C5 will have 6 spots to sit as A1, A2, B3, and B4 have already sat down.
C6 will have one spot to sit, as he/she must always be opposite to C5.
D7 will have 4 spots to sit, as 6 people have already say around the table.

and then it continues on...

 

[mchew92]

Hi, another question:

Five couples are to be seated around a dinner table. What is the probability that each member of a couple ends up opposite their partner?

Draw a circle with 10 boxes around it
allow the first Man to sit down, His wife has only one selection of a seat to sit in
for the second man...he can sit anywhere is the 8 seats remaining and his wife hence has only one selection
the thrid man then has 6 seats to choose, wife has 1 again...and so on

so ans is given by like 8 x 6 x 4 x 2 or something like tht

thts the easiest way i can think of..btw bakes told us to try avoid using Permutations button on calc.
and ill bring USB tmr for papers

 

[cutemouse]

yeah it would be a good idea to avoid using perm button.

 

[ninetypercent]

The trick to these questions is interpretation. Try to understand what they are trying to get at.
A majority of these can be solved without the PERM button.

 

[cutemouse]

therefore solution is 4C1x4C1x4C1x4C1x4C1 x no. of straights you can make

just write it out it, doesnt take too long

then divide this by 40C5

Sorry, but where did you get the '4C1' from? Arghh, I hate this topic so much!

 

[ninetypercent]

4C1 comes from the fact that with each number you choose (e.g. 1) there are 4 to choose from.

so say if you want to pick the number 1 out of the set of cards.

There are 4 ones in the set of cards: that is, red 1, blue 1, yellow 1, green 1, and you are choosing one card from the 4 cards.

and hence, the number of ways you can do this is 4C1

But you want to form a list of five numbers

so it will be 4C1 x 4C1 x 4C1 x 4C1 x 4C1

But there is no way to guarantee that this will give you five numbers in ascending order

so then you multiply it by the number of ways you can arrange the 5 numbers in ascending order

 

[Timothy.Siu]

card one...
is it just [(40x36x32x28x24)/5!]/40P5

 

[scardizzle]

card one...
is it just [(40x36x32x28x24)/5!]/40P5

how have you accounted for the ascending order part of the question?
Also since this is a card question order doesn't matter so why have you divided by 40P5

 

[Timothy.Siu]

how have you accounted for the ascending order part of the question?
Also since this is a card question order doesn't matter so why have you divided by 40P5

u talk about the ascending part and then say that order doesn't matter?
(btw, my answers probably wrong)

 

[ninetypercent]

no of ways to arrange 1,2,3,4,5,6,7,8,9,10 in ascending order. card one can only be 1,2,3,4,5 and 6

with 1 as the first card = 8 + 7 + 6 +5 +4 + 3 + 2 + 1 = 9 x 4 = 36
with 2 as first card = 7 + 6 + 5 + 4 + 3 + 2 + 1 = 8 x 3 + 4 = 28
with 3 as the first card = 6 + 5 + 4 + 3 + 2 + 1 =7 x 3 = 21
with 4 as the first card = 5 + 4 + 3 + 2 + 1 = 6 x 2 + 3 = 15
with 5 as the first card = 4 + 3 + 2 + 1 = 10
with 6 as the first card = 3 + 2 + 1 = 6

Total = 116 no of ways to make cards in ascending order

e.g. you want 1,2,3,4,5

4 ways of picking each number
4^5 ways of picking a set of 5= 1024
1024 x 116 = 118784

no restriction = 40P5 = 78960960

probability = 118784/78960960

OMG this looks wrong

EDIT: Timothy, I think u may be right.

 

[Timothy.Siu]

well, i intepreted it as the order of the cards mattered. so, 1 5 3 7 8 is different to 1 3 5 8 7 for me

 

[ascentyx]

how have you accounted for the ascending order part of the question?
Also since this is a card question order doesn't matter so why have you divided by 40P5

Card order does matter otherwise u could say 10,9,8,7,6 is ascending as it is the same as 6,7,8,9,10. Without order there is no way to determine whether it is ascending or not lol.