a How do you find the permutations of n objects taken r at a time?
B
b How do you find the combinations of n objects taken r at a time?
C
c Compare the permutations and the combinations of n objects, taking n objects at a time and taking n-1 objects at a time.
A
a _4P_4 = _4P_3, see solution.
B
b _4C_4 < _4C_3, see solution.
C
c See solution.
Practice makes perfect
a Consider a set of 4 objects. We are asked if there are more permutations of all 4 objects or of 3 of the objects. First, let's recall the formula for the permutations of n objects taken r at a time.
_nP_r = n!/( n- r)!Let's compare the permutations in a table! Remember that 0! and 1! are both equal to 1.
Permutations
n!/(n-r)!
Simplify
_4P_4
4!/( 4- 4)!
4!=24
_4P_3
4!/( 4- 3)!
4!=24
We can see that these quantities are equal!
_4P_4 = _4P_3
b Now we are asked if there are more combinations of all 4 objects or of 3 of the objects. First, let's recall the formula for the combinations of n objects taken r at a time.
_nC_r = n!/( n- r)! * r!We will compare the combinations in a table!
Combinations
n!/(n-r)!* r!
Simplify
_4C_4
4!/( 4- 4)!* 4!
4!/4!=1
_4C_3
4!/( 4- 3)!* 3!
4!/3!=4
This time, we can see that there are more combinations taking 3 objects at a time than taking 4.
_4C_4 < _4C_3
c Since both 0! and 1! are equal to 1, the permutations of n objects taken n at a time are equal to the permutations of n objects taken n-1 at a time.
_nP_n = _nP_(n-1)
On the other hand, let's look at the combinations of n objects taken n and the combinations n objects taken n-1 at a time.
Combinations
n!/(n-r)!* r!
Simplify
_nC_n
n!/( n- n)!* n!
n!/n!=1
_nC_(n-1)
n!/( n-( n-1))!* ( n-1)!
n!/(n-1)!=n
We can see that when n>1 the combinations taking n-1 objects at a time are greater than the combinations taken n objects at a time.
_nC_n < _nC_(n-1)
Note that if n=1, both quantities are the same!
_1C_1 = _1C_0
