Question

Decide if the following problem is an example of a permutation or a combination. State your rationale. For a study, 4 people are chosen at random from a group of 10 people. How many ways can this be done?
a.
combination - set of objects arranged
in a certain order
c.
combination - an unordered arrangement of objects
b.
permutation - set of objects arranged
in a certain order
d.
permutation - an unordered arrangement of objects

Answer 1

P, permutation, picking, you pick an order

C, looks like a G, combinations, you group things so order is orrelevant

Answer 2

**irrelevant

Answer 3

okay, thank you. Could you help me out with some more permutation questions?

Answer 4

maybe, just keep that rule of thumb in mind

Answer 5

How would you solve this?

Answer 6

how many groups of 1, can you make out of 325 elements?

Answer 7

Oh that's how you read that? Okay, that makes sense. So it'd be 325

Answer 8

yep, but let me dbl chk myself

spose we had 3 elements

a b c

how many groups of 1 can we make?

a
b
c

yep 3 groups of 1

Answer 9

Okay! how would you solve a more complicated one like this?

Answer 10

do you know what a factorial is?

Answer 11

Okay! how would you solve a more complicated one like this?

Answer 12

or, we can use the identity:

k! nCk = nPk

nCk = nPk/k!

k! is notation for a factorial number

Answer 13

It's where you mutiply descending numbers?

Answer 14

*multiply. For ex) 3!= 3 x 2 x 1

Answer 15

yes, and nPk means we descend for k terms

\[10C4*4C2\]
\[\frac{10P4}{4!}*\frac{4P2}{2!}\]
\[\frac{10.9.8.7}{4.3.2}*\frac{4.3}{2}\]
\[\frac{10.9.8.7}{8.3}*\frac{4.3}{2}\]
\[\frac{10.9.7}{3}*\frac{4.3}{2}\]
\[10.3.7*2.3\]

Answer 16

nPn = n!

Answer 17

oh okay, I think I get it now

Answer 18

since nCk = (nPk)/k! and k! = kPk

\[nCn=\frac{nPn}{nPn}\]
and anything divided by itself is 1, except for 0/0

so, what this simply means is that there is only 1 way to group a set n, by size n