MATH 1302 Discrete Mathematics.
Discussion Assignment Unit 2.
1. Explain in words why the equation (nk)=(n−k
n
) is true.
Solution.
The statement (nk)refers to combinations and is called “n choose k”. This means, out of a
total of n objects there are number of groups that can be defined of k quantity. For
instance, out of total eight (8) letters in the name WANYONYI, how many 2 letter
words, e.g., NO, WN, etc. meaningful or meaningless can be made, or 8 chooses 2,
written as (82).
With ()
n
k
combinations there are no repetitions and order does not matter. Therefore, let
us consider an example: ( )
50
4
50! = 50*49*48*47*…….
50! = 50*49*48!
This tells us that 48! Is part of the 50! hence implicitly in in the numerator of the
combination formula and cancels out the k! in the denominator part of the formula.
2. Use the definition (nk ) = k ! ( n−k
n!
)!
to show that the equation in question 1 is true.
Solution.
Substitute k with (n-k):
(n−k
n =
) ( n−k ) ! ( n−
n!
( n−k ) ) !
(n−k
n =
) ( n−k ) ! (nn−n+k
!
)!
(n−k
n =
) ( n−kn!) ! ( k ) !
Rearranging the equation gives us:
1|Brian Wanyonyi – University of the People.
Discussion Assignment Unit 2.
1. Explain in words why the equation (nk)=(n−k
n
) is true.
Solution.
The statement (nk)refers to combinations and is called “n choose k”. This means, out of a
total of n objects there are number of groups that can be defined of k quantity. For
instance, out of total eight (8) letters in the name WANYONYI, how many 2 letter
words, e.g., NO, WN, etc. meaningful or meaningless can be made, or 8 chooses 2,
written as (82).
With ()
n
k
combinations there are no repetitions and order does not matter. Therefore, let
us consider an example: ( )
50
4
50! = 50*49*48*47*…….
50! = 50*49*48!
This tells us that 48! Is part of the 50! hence implicitly in in the numerator of the
combination formula and cancels out the k! in the denominator part of the formula.
2. Use the definition (nk ) = k ! ( n−k
n!
)!
to show that the equation in question 1 is true.
Solution.
Substitute k with (n-k):
(n−k
n =
) ( n−k ) ! ( n−
n!
( n−k ) ) !
(n−k
n =
) ( n−k ) ! (nn−n+k
!
)!
(n−k
n =
) ( n−kn!) ! ( k ) !
Rearranging the equation gives us:
1|Brian Wanyonyi – University of the People.
