Permutations Help (1 Viewer)

[academical]

Just a few questions I would realy appreciate some help with (if you could explain how you got the answers, that would be great!)...

1. A table has 4 boys and 4 girls sitting around it.

a) If the seating is arranged at random, find the probability that
i) 2 particular girls will sit together.
ii) all the boys will sit together.

2. Nine beads are arranged randomly in a line.
a) The beads are arranged on a bracelet. How many different ways are possible?

3. Find the probability that in a cirlce of 20 people, 2 particular people will be together.

4. How many different arrangements can be made from the following words? Note that the arrangements need not be proper words.
a) ENGINEERING
b) SUPERMARKET

Thank you in advance for your help

 

[zeebobDD]

2. there are nine beads and arranged randomly on a bracelet (circular) so it's 8! since circular objects are all (n-1)!

 

[Aesytic]

1. a) assuming the table is round,
if you group the 2 particular girls together as 1, then you'll have 7 "people" (4 boys 2 girls and the group of 2 particular girls)
since it is a round table, the number of arrangements possible will be (7-1)!*2, since the girls in the group can switch positions
the overall number of arrangements with no restrictions is (8-1)! so the probability will be {6!*2}/7!
b)using a similar method, if all the boys are grouped as 1, then you'll have 5 "people"
.'. number of arrangements would be (5-1)!*4! since the boys can have 4! different arrangements within their group
.'. probability is {4!4!}/7!

2. a) 9 beads in a bracelet will be (9-1)!/2 since it is a circle, but a bracelet can be flipped over, unlike a table, and hence those arrangements which are symmetrical are not counted

3. group the 2 people together as 1, so you'll have 19 "people"
.'. number of arrangements would be (19-1)!*2 since the 2 people can have 2 different arrangements within themselves
number of arrangements with no restrictions are (20-1)!
.'. probability is {18!*2}/19!

4.a)ENGINEERING has 11 letters in it, but there are 3 E's, 3 N's, 2 G's and 2 I's
.'. number of different words will be 11!/{3!3!2!2!} to account for the fact that the letters are all the same, i.e. they are indistinguishable
b) SUPERMARKET has 11 letters in it, with 2 E's and 2 R's
.'. using a similar method number of different words will be 11!/{2!2!}

 

[academical]

2. there are nine beads and arranged randomly on a bracelet (circular) so it's 8! since circular objects are all (n-1)!

That's what I thought, but the answers say it is in fact 20,160.
8! = 40,320 (which is double the given answer).

So then I thought that it was a line (depending on how the bracelet was attached), however 9! = 362,880.

Are the answers incorrect or have I missed something?

Thanks for your help though

 

[academical]

1. a) assuming the table is round,
if you group the 2 particular girls together as 1, then you'll have 7 "people" (4 boys 2 girls and the group of 2 particular girls)
since it is a round table, the number of arrangements possible will be (7-1)!*2, since the girls in the group can switch positions
the overall number of arrangements with no restrictions is (8-1)! so the probability will be {6!*2}/7!
b)using a similar method, if all the boys are grouped as 1, then you'll have 5 "people"
.'. number of arrangements would be (5-1)!*4! since the boys can have 4! different arrangements within their group
.'. probability is {4!4!}/7!

2. a) 9 beads in a bracelet will be (9-1)!/2 since it is a circle, but a bracelet can be flipped over, unlike a table, and hence those arrangements which are symmetrical are not counted

3. group the 2 people together as 1, so you'll have 19 "people"
.'. number of arrangements would be (19-1)!*2 since the 2 people can have 2 different arrangements within themselves
number of arrangements with no restrictions are (20-1)!
.'. probability is {18!*2}/19!

4.a)ENGINEERING has 11 letters in it, but there are 3 E's, 3 N's, 2 G's and 2 I's
.'. number of different words will be 11!/{3!3!2!2!} to account for the fact that the letters are all the same, i.e. they are indistinguishable
b) SUPERMARKET has 11 letters in it, with 2 E's and 2 R's
.'. using a similar method number of different words will be 11!/{2!2!}

Thank you so much!

I really appreciate it