Question

how many permutations are there of 3 things taken from a set of 7?

Answer 1

@amistre64

Answer 2

if you are using a calculator.. you can just plug in the values 7P3

Answer 3

the way i remember the is to start at 7 and pick 3 numbers to multiply:

7*6*5 = 7P3

Answer 4

where does the 6 come from?

Answer 5

7 ways to pick the first element
6 ways to pick the second element
5 ways to pick the third element

Answer 6

if we made a tree diagram of this, it would be 7 branches with 6 branches each; 7*6

and each branch has 5 leaves: (7*6)*5 total outcomes

Answer 7

for any given set of elements, there are 6 ways that they can be presented:

abc acb bac bca cab cba, which means that for combinations, we have 6 like sets.

(7P3)/3! would be the combinatorial calculation ... but thats not really what it asked for is it :)

Answer 8

so confused. whats the simplest way to solve this?

Answer 9

start at 7 and pick 3 numbers to multiply: 7*6*5

Answer 10

so the answer is 210? i dont get how

Answer 11

how do you pick the 3 numbers?

Answer 12

i explained how ... which you say is confusing

6*5 = 30, 30*7 = 210

Answer 13

start at 7 and go down

Answer 14

spose you have a litter of 7 kittens; how many ways can you choose the first one to pick up? after picking up the first one, how many ways are left in which you can choose to pick up the next one? after youve picked 2 kittens, how many ways are left for you to be able to choose the last kitten?

Answer 15

so if it was:
how many permutations are there of 2 things taken from a set of 5?
the answer would be:
5*4
=20
?

Answer 16

7 to choose from; then 6 to choose from, then 5 to choose from

Answer 17

yes 5*4 is the permutations of 2 things from 5

Answer 18

ohh thank youu

Answer 19

youre welcome

Answer 20

"If K objects out of a set of N elements are placed into a fixed order, there are ________ unique arrangements of those objects" ??

Answer 21

start with n and start picking out k elements

or rather, theres some formula you can use ....\[\frac{n!}{(n-k)!}\]

Answer 22

huhh??

Answer 23

Are the N elements all different?

Answer 24

a set is properly defined by distinct elements ....

Answer 25

a set that has N elements is by definition N different elements

Answer 26

so the blank would be N, correct?

Answer 27

no

Answer 28

7 Pick 3 did not amount to 7

Answer 29

\[\frac{7*6*5*4*3*2*1}{4*3*2*1}=\frac{7!}{(7-3)!}\]

Answer 30

i dont get the bottom..

Answer 31

k elements picked from a set of n elements is usually calculated as: n!/(n-k)!

Answer 32

notice that in order to simplify this to 7*6*5, we have to divide out the 4*3*2*1

Answer 33

in general; to simplify n! to account for nPk, we need to divide out (n-k)!

Answer 34

ok

Answer 35

so....

Answer 36

soo lost...