stupid permutations (1 Viewer)

[pLuvia]

can some one help me do these questions

How many numbers of 7 digits can be formed from the digits 1,2,3,4,5,6,7 if:
(i) the number begins with 1 and ends with 2?
(ii) there are not more than 2 digits between 1 and 2?

Find the number of arrangements of the letters in the word PENCILS if
(i) E is next to I
(ii) E precedes I
(iii) there are three letters between E and I

A car can hold 3 people in the front seat and 4 in the back seat. In how many ways can 7 people be seated in the car if 2 particular people must sit in the back seat and 1 particular person is the driver?

 

[haboozin]

while ur doign them for kadil answer my question too:


Out of 25 students who study drama, 3 are chosen to be in a play.

A brother and a sister both do drama, find the probability that neither of them will be selected for the play.

 

[pLuvia]

while ur doign them for kadil answer my question too:


Out of 25 students who study drama, 3 are chosen to be in a play.

A brother and a sister both do drama, find the probability that neither of them will be selected for the play.

P(chosen for play) = 3/25
P(not chosen for play) = 22/25

P(bro and sis not getting chosen) = 22/25
???

 

[haboozin]

P(chosen for play) = 3/25
P(not chosen for play) = 22/25

P(bro and sis not getting chosen) = 22/25
???

sorry no
.................

i dont want to say the answer because u could easily make an equation with factorials to get it.

 

[KFunk]

while ur doign them for kadil answer my question too:


Out of 25 students who study drama, 3 are chosen to be in a play.

A brother and a sister both do drama, find the probability that neither of them will be selected for the play.

if you remove the two of them then there are ²³C₃ ways to choose three from the rest. With them included there are ²⁵C₃ ways. I'm still fairly poor at the perms/combs/binomial stuff but I think it'd be:

(²³C₃)/(²⁵C₃) = 0.77

 

[haboozin]

if you remove the two of them then there are ²³C₃ ways to choose three from the rest. With them included there are ²⁵C₃ ways. I'm still fairly poor at the perms/combs/binomial stuff but I think it'd be:

(²³C₃)/(²⁵C₃) = 0.77

cheers
78910

 

[pLuvia]

err.. can you do my ones please?

 

[KFunk]

A car can hold 3 people in the front seat and 4 in the back seat. In how many ways can 7 people be seated in the car if 2 particular people must sit in the back seat and 1 particular person is the driver?

Only one person can drive so remove him from the equation and leave him in place. Of the 6 people that are left, 4 can sit in the front so when filling those places you multiply (4 x 3). Considering the back you then have 4 people who can sit in any order yielding 4!. Hence, combining this, the answer is:

4 . 3 . 4! = 288 ways

 

[grendel]

Assuming you can only use each digit once:

(i) 5!
(ii) arrangements where they are together=6! x 2
arrangements where they are seperated by 1 digit=5 x 2 x 5!
arrangements where they are seperated by 2 digits =4 x 2 x 5!
tf total arrangements = 3600

(i) arrangements where e is next to i=6! x 2
(ii) 6!
(iii) 5! x 2 x 3

if you need further explanation, post back.

 

[KFunk]

Find the number of arrangements of the letters in the word PENCILS if
(i) E is next to I
(ii) E precedes I
(iii) there are three letters between E and I

i) consider E and I as a single unit such that you are dealing with 6 individual units (instead of 7). There are 6! ways of arranging these units but 2 ways of arranging E and I between themselves so the answer becomes:

2 . 6! = 1440

ii) This is an annoying one to explain in full but the way I worked it is to consider all the possible cases and add them together i.e E is in the 1st place, the 2nd place ... through to the 6th place. The number of letters which can go before E includes all the letters except for I. The number of letters which can go after includes E. So the different cases are:

E first = (E) 6!
E second = 5 . (E) . 5!
E third = 5 . 4 . (E) . 4!
E fourth = 5 . 4 . 3 . (E) . 3!
E fifth = 5! (E)

Adding these together yields --> 2,520 arrangements

iii) If you consider that E is in the first place and I in the fifth then you have 5! ways of arranging the other letters and 2 ways of arranging E and I amongst themselves. This case is the same the case where E is in the second place or the third place (the other possibilities) so you can just multiply through by 3 giving:

3 . ( 5! . 2 ) = 720 arrangements

 

[pLuvia]

A committee of 7 politicians is chosen from 10 liberal members, 8 labor members and 5 independents. In how many ways can this be done so as to include exactly 1 independent and at least 3 liberal memebers and at least 1 labor member?

can someone help me do this?

 

[pLuvia]

but its suppose to be AT LEAST 3 liberal and AT LEAST 1 labor, doesnt that change the working out of it?

 

[pLuvia]

but like the answer is 73080

 

[pLuvia]

1 must be independent ...[1]
1 must be labor ...[2]
3 must be liberal ...[3]
2 can be either labor or liberal but cannot be independent ...[4]

[1] can happen in ⁵C₁ different ways
[2] can happen in ⁸C₁ different ways
[3] can happen in ¹⁰C₃ different ways
[4] can happen in ¹⁴C₂ different ways
(since there are now 7 liberal and 7 labor left)

So these 4 (successive) events can happen in ⁵C₁ x ⁸C₁ x ¹⁰C₃ x ¹⁴C₂ different ways

i dont understand how you got it lol, im having trouble understanding permutations and combinations

 

[SaHbEeWaH]

So these 4 (successive) events can happen in 5C1 x 8C1 x 10C3 x 14C2 different ways

This is impossible.. choosing 7 from 23 is ²³C₇ without restrictions, so it cannot be higher than this.

You have to take each possibility one at a time:

1 independent, 1 labor, 5 liberal
1 independent, 2 labor, 4 liberal
1 independent, 3 labor, 3 liberal

(⁵C₁ x ⁸C₁ x ¹⁰C₅) + (⁵C₁ x ⁸C₂ x ¹⁰C₄) + (⁵C₁ x ⁸C₃ x ¹⁰C₃)

 

[pLuvia]

Combination Question

Combination Question

14 The number plates of a motor car contain 3 letters of the alphabet followed by 3 numerals. What proportion of these would contain 3 letters the same and 3 numerals the same?

15 An urn contains 5 red cubes and 4 white cubes. Three cubes are drawn in succession without replacement. What is the probability that
(a) the first two cubes are red and the third one white.

 

[KFunk]

Combination Question
The number plates of a motor car contain 3 letters of the alphabet followed by 3 numerals. What proportion of these would contain 3 letters the same and 3 numerals the same?

Number of unique plates = (26)³.(10)³ (3 options of 26 letters each and 3 options of 10 digits)

Number of Plates with 3 letters same and 3 numerals the same = 26.10 =260 (since there are 26 letters that can be repeated thrice and 10 digits to be repeated thrice)

The proportion = [260]/[26³.10³] = 1/67,600

 

[KFunk]

Combination Question

15 An urn contains 5 red cubes and 4 white cubes. Three cubes are drawn in succession without replacement. What is the probability that
(a) the first two cubes are red and the third one white.[/B]

You can do this with basic probaility:

1^st draw --> P(red) = 5/9

2^nd draw --> P(red) = 4/8

3^rd draw --> P(white) = 4/7

P(red, red, white) = (5.4.4)/(9.8.7) = 10/63

 

[pLuvia]

thanks KFunk~~

 

[pLuvia]

Eight people are to be divided into two groups. What is the probability there will be 4 in each group