PErms and comb (1 Viewer)

[GaganDeep]

How many numbers greater than 4000 can be formed from the figures 3,5,7,8,9? repitition not allowed?

How many 7 digits can be formed from the digits 1,2,3,4,5,6,7 if there are no more than 2 digits between 1 and 2

Find the number of arrangements of letters of pencils if there are three letters between e and i

In how many ways can 6 people be seated in a motor car if only 2 can occupy the drivers position?

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 particlar person is the driver

In how many ways ways can 4 people be accommodated if there are 4 rooms available?

In how many ways can 5 men and 5 women by arranged in a circle so that the men are seperated? In how many ways can this be done if two particular women must not be next to a particular man?


How many different slections can be made by taking 3 of the digits 4,5,6,7,8,9?
why do u use combinations, why not use perms.


In how many ways can n things be shared between 2 people

 

[Riviet]

How many different slections can be made by taking 3 of the digits 4,5,6,7,8,9?
why do u use combinations, why not use perms.

There are 6 numbers available, we choose 3 of them, therefore there are 6C3 ways. We use combinations because choosing 4 5 6 is the same as choosing 4 6 5 and 6 5 4, etc. in other words, order doesn't matter.

 

[shimmerz_777]

1

2 cases, you use either all 5 numbers, 5! ways or only 4
in the case of 4 you have 4 options for ur first number ( needs to be greater than 4) and the rest can be anything so its 4x4x3x2
add them together and its 216 ways

2
if 2 or 1 occupy the first position, then the amount of combinations is

2x5x4x1x3x2x1
this can be translated so that 2 or one is in the second position and 3rd... 5 times so the answer is 5x2x4x1x3x2x1, = 1200 EDIT---( forgot to account for when they are next to eachother or seperated by one, see webby's answer)

3
same way as above, its 3 ( 2x5x4x3x1x2x1) = 720 ways

4 i assume the car has 6 seats, in this question its almost 6!, but in the instance where u have 6 people whou could take the saet, you can only have 2, so its 2x5x4x3x2x1, = 240

5 since one has to be the driver, take him out of the equasion. 2 seats are left in the front in which 2 ppl cannot occupy, so there are 4x3 ways in which they can be taken, now the back seat isnt restricted anymore so its 4! ways
answer = 4x3x4!= 288

6) if they occupy one room each its 4!

7) hard to explain. if they are seperated they have to be inbetween women, in which case the seating arangements in a circle are 5x5x4x4x3x3x2x2x1x1/10 as there are 5 men for the first seat, and 5 women can occupy the seat next to him, then 4 men for the next and so on, and since its a circle u must divide by 10, giving 1440 ways

8) riviet answered

9) this one is hard, and im not entirely sure. my thoughts are there are n+2 amounts of divisions, each has nCr ways, r being how many things one person has. timesing them all together is confusing and i think i have done enough for now

PS this is my best 3u topic, people find it hard but for some reason i dont. all my methods were done doodleing on a piece of paper or in my head so i cant give a method for it

 

[shimmerz_777]

ha webby i did all mine at once so i didnt see any of ur answers, plus u forgot to multiply by 2 in your driver seat one, EDIT--i mean u wrote it down but u didnt calculate it

 

[shimmerz_777]

for the last question have u concidered the different ways in which someone can hold am amount of objects? your method is a bit vague, i mean if someone has 5 objects and the other has (n-5) then there are nC5 ways in which this can occur. or am i just over thinking

 

[shimmerz_777]

very well done, its just that ur method was 7 words and i couldnt really follow the idea behind it