Additional Problem Set 5 Solutions

1
.
The door to a building has a lock which has 5 buttons numbered from 1 to 5. The combinations of numbers that opens the lock is a sequence of 5 numbers and is reset every week.
1
.
1
.
This is just a permutation of 5 elements 5!
5! = 120
1
.
2
.
Choose the buttons that are going to be pressed together at the beginning C(5,2)Considering number of ways of arranging the remaining numbers 3!​

5
2
3!=60
2
.
Choose the processors that are going to be used C(3,2) = 3
Choose between the two processors where to assign the n tasks
n
2
. Then take out the cases in which all processes get assigned to one processor -2
3
(
n
2
-2)
3
.

10
5
=252
4
.
Binomial[n,k] = Binomial[n-1,k-1] + Binomial[n-1,k] We will show two proofs
4
.
1
.
The left hand side can be considered as the way of choosing k elements out of a set of n. The right hand side can be seen as the way of doing the same by two cases. Case 1. When a particular element is in the set. Then we have to choose the remaining k-1 from the remaining n-1. Case 2. When the particular element is not in the set, so we have to choose the k elements out of the n-1 other elements.
4
.
2
.
n=nn= k+n-k(n-1)! n = (n-1)!k + (n-1)! (n-k)n! = (n-1)!k + (n-1)! (n-k)​
n!
k!(n-k)!
=
(n-1)!k
k!(n-k)!
+
(n-1)!(n-k)
k!(n-k)!
​​
n!
k!(n-k)!
=
(n-1)!
(k-1)!(n-k)!
+
(n-1)!
k!(n-k-1)!
​​

n
k

=

n-1
k-1
+
n-1
k

​​
5
.
We will show two ways of counting
5
.
1
.
Choose the two colors to use C(3,2)For each nail then you have 2 possibilities depending on the color
5
2
​But we have counted all red nails, all blue nails, and all yellow twice each, so we subtract 3.​

3
2

5
2
-3=93
5
.
2
.
Count the number of ways to color all nails the same color C(3,1) = 3Count the number of ways to color the nails with exactly 2 colors We first choose the Colors C(3,2)=3 Then we consider the choice for each of the five nails
5
2
but subtract the case when all nails were colored the same (-2)3 +
3(
5
2
-2)
=
93