Question

I need some help with Permutations and Combinations. Someone please help me.

Answer 1

Question: There are 4 identical green marbles and 3 identical blue marbles arranged in a row.
a) How many different arrangements are possible?
b) How many different arrangements of just five of these marbles are possible?

Answer 2

@bigbang123 Did you try to do it first?

Answer 3

I'm really having troubles with Permutations and Combinations, I'm terrible at english. If you could help explain. I would be really happy.

Answer 4

Why not.. But first let me ask a last off topic question , is it your test question ?

Answer 5

@mathslover No. It's my homework question. Why?

Answer 6

Nothing , just we are not allowed to answer test questions as it misguides the users. No problem. I will help you

Answer 7

@mathslover Thank you sir.

Answer 8

See total arrangments possible are : n_P_r / 3!*4!

Answer 9

that is : 7_P_7 / 3! *4!

Answer 10

@mathslover can you explain how do you get that answer?

Answer 11

yes why not? See first of all it is clear that the req. answer will be in the form of :

n_p_r / 3! *4!

now since total arrangements possible = 7 and req. arrangements = 7

thus n = r = 7

7_P_7 / 3! *4!

Answer 12

ok.

Answer 13

case - 1 :

4 green + 1 blue

total no. of ways = 5_P_5 / (4! * 1! )

case - 2

3 green + 2 blue

total no. of ways = 5_P_5/ (3! * 2!)

case - 3

2 green + 3 blue

total no. of ways = 5_P_5 / (2! * 3!)

Answer will be all the sum of the total no. of ways in all the cases.

Answer 14

^ Second one

Answer 15

can you explain more in dept please? im really confused. im failing right now :(

Answer 16

sorry I have to go , you can learn it here :
http://www.mathsisfun.com/combinatorics/combinations-permutations.html

Answer 17

thanks

Answer 18

if you could tell the marbles apart, it would be 7! but since you cannot tell the 4 green apart or the 3 blue apart it is \(\frac{7!}{4!3!}\)

Answer 19

ok

Answer 20

Why are you bumping this? Why don't you close it?