Each of these is the samecombinationof the objects A, B, C – that is a selection of
three objects in which order is not important.
Now suppose we select justrobjects from then. Each such selection is a different
combination ofrobjects fromn. An obvious question is how many different permutations
ofrobjects chosen fromncan be formed in this way? This number is denoted bynPr.
It may be evaluated by repeating the previous counting procedure, but only until we have
chosenrobjects:

The first may be chosen innways

The second may be chosen in (n−1) ways

The third may be chosen in (n−2) ways

..

.

Therth may be chosen in (n−(r− 1 ))ways

So the total number of permutations will be

nP

r=n×(n−^1 )×(n−^2 )×...×(n−r+^1 )

=

n(n− 1 )(n− 2 )...(n−r+ 1 )(n−r)(n−r− 1 )... 2 × 1

(n−r)(n−r− 1 )... 2 × 1

=

n!

(n−r)!

For example the number of ways that we can permute 3 objects chosen from 5 distinct
objects is

(^5) P
3 =^5 ×^4 ×^3 =
5 × 4 × 3 × 2 × 1
2 × 1

5!
( 5 − 3 )!
= 60
Since the order does not matter in acombination,nPr will includer! permutations
of the same combinations ofrdifferent objects. So thenumber of combinations ofr
objects chosen fromnis
1
r!
nP
r=
n!
(n−r)!r!
This is usually denoted bynCr(called the ‘n−C−r’ notation) or
(
n
r
)

• ‘chooser

objects fromn’:

nCr=

(

n

r

)

=

n!

(n−r)!r!

e.g.^5 C 3 =

(

5

3

)

=

5!

( 5 − 3 )!3!

= 10

which is very useful in binomial expansions (see Section 2.2.13) and other areas, simply
as a notation, regardless of its ‘counting’ significance.