Permutations (1 Viewer)

[tempco]

I need some help with a few permutation/combination questions.

---
(A)

A die is loaded in such a way that in 8 throws of the die, the probability of getting 3 even numbers is four times the probability of getting 2 even numbers.

Find the probability that a single throw of the die results in an even number.

---
(B)

There are five gentlemen and four ladies to dine at a round table. In how many ways can they seat themselves so that no two ladies are together?

---
(C)

Mary is to celebrate her 18th birthday by having a dinner party for herself and 11 other people. Mary is to sit at the head of the table.

If there are six men and six women at the party, and Mary decides to seat the men and women alternatively, in how many ways can this be done?

---
(D)

Seven people enter one compartment of a railway carriage in which there are 10 seats, 5 facing forward and 5 facing the rear. In how many ways can they take their sears if two particular people (say A and B) sit together facing the front.

---

help is much appreciated... thanks!

 

[withoutaface]

A) This is binomials I think, although it's working backwards. You go:

let P(even)=p

:. 8C3(p)³(1-p)⁵=4*8C2(p)²(1-p)⁶

and you solve it for p.

B) You start by seating two men together, and they have 5P2 ways to do this.

with the remaining 3 men you have 3 spots, hence 3!

with the four women you have 4 spots, hence 4!

answer is 5P2*3!*4!

still working on the others

EDIT:

C) the answer is 5!*6! because Mary is already seated, so her place is already determined, and the rest of the women have five spots, while the 6 men have six spots.

D) 5P2*5P5+5C1*5P3*5P4+5C2*5P4*5P3+5C3*5P5*5P2

As always, I'd check these answers with what's in the book, because one or more may be wrong.

 

[ngai]

I need some help with a few permutation/combination questions.

by inspection, and checked by counting every single possibility:
(A) P = 2/3
(B) 2880 ways
(C) 86400 ways
(D) 53760 ways

 

[withoutaface]

Ngai can you please explain the last one to me, because I just can't see it.

 

[tempco]

Thanks for the replies... time to figure them out

 

[Xayma]

Ngai can you please explain the last one to me, because I just can't see it.

Come on that is a MIF question

You have the seats as following:

__ __ __ __ __

__ __ __ __ __

Now two people A and B sit together in the bottom row.
The left most person can sit in one of 4 seats and there are 2! ways they can sit between themselves.

Ie there are 8 ways these two people can sit together.

Then there are 5 other people with 8 other seats.

Ie ⁸P₅

Therefore the total number of ways is:

⁸P₅*8=53 760

 

[tempco]

Refering to (B), would the answer be the same if the condition was that Mr A and Mr B had to sit together, without any more conditions like alternate sex sitting?