OR360
Problem Set 1
Solutions to Problems 1 and 2
1.
Use the definition of the “choose” notation and compute directly, starting from the righthand
side. Bring the two fractions to a common denominator so they can be added, then simplify:
n

1
r

1
+
n

1
r
=
(
n

1)!
(
r

1)!(
n

r
)!
+
(
n

1)!
r
!(
n

r

1)!
=
(
n

1)!
r
+ (
n

1)!(
n

r
)
r
!(
n

r
)!
=
(
n

1)!
n
r
!(
n

r
)!
=
n
!
r
!(
n

r
)!
=
n
r
For a combinatorial argument, consider a particular element
X
in a group of
n
elements.
A
subset of
r
elements out of this group either contains
X
or it does not, hence the total number
of subsets equals the sum of the number of subsets containing
X
and the number not containing
X
. In order to choose a subset containing
X
, we take
X
into the subset, and choose another
r

1 elements out of the remaining
n

1 elements in the group, so there are
(
n

1
r

1
)
ways to do
this. In order to choose a subset
not
containing
X
, we place
X
aside and choose the
r
elements
out of the remaining
n

1 elements, so there are
(
n

1
r
)
ways to do this.
2.
There are many different ways to prove this. The easiest to note is that if
A, B, C
are indepen
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 Summer '08
 WEBER
 Lefthandedness, n1 ways, n1 + r1

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