MATH 109
Pascal’s Triangle
Pascal’s Triangle gives all the ways to choose subsets of objects from a group of N
distinct objects. For example, from a group of N = 4 objects, we can choose none (in 1
way), we can choose one object (4 ways), we can choose two objects ( C(4, 2) = 6 ways),
we can choose three objects ( C(4, 3) = 4 ways), or we can choose all four objects (in 1
way).
To obtain a number in the interior, simply add the two numbers above it.
1
1
1
1
1
C(0, 0)
1
2
3
C(1, 0) C(1, 1)
1
3
C(2, 0) C(2, 1) C(2, 2)
1
4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
C(3, 0) C(3, 1) C(3, 2) C(3, 3)
C(4, 0) C(4, 1) C(4, 2) C(4, 3) C(4, 4)
C(5, 0) C(5, 1) C(5, 2) C(5, 3) C(5, 4) C(5, 5)
C(6, 0) C(6, 1) C(6, 2) C(6, 3) C(6, 4) C(6, 5) C(6, 6)
etc.
Note the symmetry in the triangle. From a group of N = 4 objects, if we choose 1 to
include, then it is the same as choosing 3 to exclude. So C(4, 1) = C(4, 3) = 4.
Likewise, from a group of N = 7 objects, if we choose 2 to include, then it is the
same as choosing 4 to exclude. So C(7, 2) = C(7, 5) = 21. If we choose 3 to include,
then it is the same as choosing 4 to exclude. So C(7, 3) = C(7, 4) = 35.
Note also that C(N , 0) = C(N , N ) = 1 because there is only 1 way to select none and
only 1 way to select all. Moreover, C(N , 1) = N because there are N ways to select 1
object from a group of N objects.
Example 1. From a group of 7 objects, how many ways are there to select anywhere
from 0 to 2 objects? How many ways are there to select anywhere from 3 to 6 objects?
Solution. In the first case, C(7, 0) + C(7, 1) + C(7, 2) = 1 + 7 + 21 = 29. In the second
case, C(7, 3) + C(7, 4) + C(7, 5) + C(7, 6) = 35 + 35 + 21 + 7 = 98.
Exercises
1. From 6 objects, how many ways are there to select anywhere from 0 to 3?
2. From 5 objects, how many ways are there to select anywhere from 2 to 4?
Subsets from a Group of Size N
If we have a set of size N , then there are various subsets we can create. A subset can
have no elements (i.e., it is the empty set ! ); it can have one element (we choose 1), it
can have two elements (we choose 2), etc. If we want a subset of size k , then we must
choose k elements. So there are C(N , k) possible subsets of size k from a set of size N .
The total number of all possible subsets of any size would be C(N , 0) + C(N , 1) +
C(N , 2) + . . . + C(N , N ) (i.e., add up an entire row of Pascal’s Triangle.) This row will
N
add up to 2 . To create an arbitrary subset, there are two choices for each of the N
elements “In” or “Out.” So the total number of possible subsets of any size is
N
2 !2!... !2 = 2 .
Example 2. The soccer team has 14 players. (a) How many subsets of size 9 can be
created? (b) How many subsets (of any size) are there?
Solution. (a) We simply choose 9 from 14 giving C(14, 9) = 2,002 possibilities. (b) There
14
are 2
= 16,384 subsets altogether ranging from size 0 to size 14.
Exercises
3. From a group of 7 objects, how many ways are there to select an arbitrary subset (i.e.,
anywhere from none to possibly all)?
4. The little league basketball team has 12 players. (a) How many possible subsets of
any size are there of the whole team? (b) How many subsets of size 5 are there? (c)
Among the 5 starters, how many possible subsets are there?
5. A diner is ordering a chef salad. Available are 3 types of lettuce, 5 types of meat, 8
types of vegetables, 6 types of dressing, and 4 toppings. How many possibilities are
there if one is allowed to choose
1 type of lettuce, from 0 to 2 types of meat, from 0 to 8 types of vegetable,
1 type of dressing, and from 0 to 3 types of topping?
Answers
1. C(6, 0) + C(6, 1) + C(6, 2) + C(6, 3) = 1 + 6 + 15 + 20 = 42.
2. C(5, 2) + C(5, 3) + C(5, 4) = 10 + 10 + 5 = 25.
3. Sum the entire row: C(7, 0) + C(7, 1) + C(7, 2) + . . . . + C(7, 7) = 128. Alternately,
for each of the seven objects, make one of two choices Yes/No. The total number of
7
choices becomes 2 ! 2 ! 2 ! 2 ! 2 ! 2 ! 2 = 2 = 128.
12
4. (a) 2
= 4096
(b) C(12, 5) = 792
5
(c) 2 = 32
5. 3 ! (1 + 5 + 10) ! 2 8 ! 6 ! (1 + 4 + 6 + 4) = 3 ! 16 ! 256 ! 6 ! 15 = 1,105,920