Some problems involving nCk
MAD 2104
(Solutions below)
1)How many diagonals does an hexagon have?
2)How many binary strings of length n contain exactly k 1s ?
3)How many length-10 strings with 0-1-2-3-4 contain exactly three 1s?
4)How many length-10 strings with 0-1-2-3-4 contain exactly three 1s and five 0s?
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5)With n tennis players,
a) what is the number of “singles” (matches) that can be set up?
b) what is the number of “doubles” that can be set up?
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6)There are two types of fellowships (A and B) to be awarded, and the final screening results in 8
candidates. According to the regulations each fellowship will be awarded to one or two candidates. In
addition, no candidate can receive more than one fellowship. In how many ways can this be done?
7)The twelve hours in the clock face are to be colored. In how many ways can this be done if
a) 6 of them must be white, 4 must be black and 2 must be red?
b) 6 of them must be white, the others may be blue or red?
c) the even hours are white, the odd ones alternate between blue and red?
8)Prove that the product of n consecutive natural numbers is always a multiple of n!
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9) Eight people are standing on the ground floor of a building, waiting for the elevator. If the elevator can
carry up to 5 people, in how many ways can it leave the ground floor? (Note: the order in which people
enter the elevator does not matter)
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10) Problems involving teams. Notice that the ordering is irrelevant (unless otherwise specified). Hence
{a,b} as a team is equal to {b,a}, where a and b are persons.
All these problems refer to a group G of twelve people, that is, G is the “pool”.
I)
a)
How many five-person teams can be chosen from G?
b)
Suppose two members (A and B) of G insist on working as a pair. That is, the teams must contain
both of them or neither of them? How many five-person teams can now be chosen?
c)
On the contrary, supposed that A and B do not get along and so refuse to work together on a team.
How many five-person teams can now be formed?
II)
Let us now be more specific: we know that G contains five men and seven women. Use
this additional information to answer these questions.
d)
How many five-person teams can be chosen consisting of three men and two women?
e)
How many five-person teams consist of at least one man?
f)
How many five-person teams contain at most one man?
Recall: at least one, means 1 or more. At most one means 0 or 1.
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Some problems involving nCk
MAD 2104
SOLUTIONS
1) How many diagonals does an hexagon have?
Answer: each pair of vertices determine a chord: hence there are 6C2 chords, that is, 15. These include the
six sides (which are not diagonals). Therefore, the number of diagonals is 15 – 6 = 9.
2) How many binary strings of length n contain exactly k 1s ?
Answer: Task 1: choose k boxes where to place the 1s; hence, in nCk ways.
Task 2: fill the remaining n-k boxes with zeros; in just one way. So the answer (by the FCP) is nCk .
3) How many length-10 strings with 0-1-2-3-4 contain exactly three 1s?
Answer: Task 1: choose where to place the 1s. In 10C3 ways. Task 2: fill the remaining 7 boxes with digits
from the pool {0,2,3,4}. In each box we have 4 possibilities, so for task 2 we have 47 ways.
By the FCP the total number is (10C3 ). (47) = 393216
4) How many length-10 strings with 0-1-2-3-4 contain exactly three 1s and five 0s?
Answer: Task 1: choose where to place the 1s. In 10C3 ways.
Task 2: Choose where to place the 5 zeros. Since there are only 7 empty boxes, we do this in 7C5 ways.
Task 3: fill the two remaining boxes with numbers from the pool {2,3,4}. In 32 ways.
By the FCP, the total number is 10C3 . 7C5 .9 = 22680
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5) With n tennis players,
a) what is the number of “singles” (matches) that can be set up?
Answer: we need two players to define each match. There are nC2 possibilities.
b) what is the number of “doubles” that can be set up?
Answer: Task 1: choose 4 players, in nC4 ways.
Task 2: set up the doubles once the four players have been chosen. In three ways, because if we call the
players
a,
b,
c
and
d,
the
games
are
{a,b}
vs
{c,d},
{a,c} vs {b,d}, and {a,d} vs {c,b}.
By the FCP, the answer is (3). (nC4 )
6) There are two types of fellowships (A and B) to be awarded, and the final screening results in 8
candidates. According to the regulations each fellowship will be awarded to one or two candidates. In
addition, no candidate can receive more than one fellowship. In how many ways can this be done?
Answer: we have to consider the different possibilities, which we divide into four cases. In each case we
will resort to tasks and FCP. Details are left to the reader!
CASE 1: A awarded to one candidate, B likewise. This can be done in 8.7=56 ways
CASE 2: A awarded to one candidate, B to two. This can be done in
(8). (7C2) ways.That is, 168 ways.
CASE 3: A awarded to two candidates, B to one. This is similar to case 2, so can be done in exactly 168
ways.
CASE 4: A awarded to two candidates, B likewise. This can be done in (8C2). (6C2) ways, that is, 420
ways.
Having analyzed all the cases, we have to add (why?)the numbers, so the answer is 56+168+168+420= 812
ways.
Remark. Please notice where we used the fact that no candidate can receive more than one fellowship.
7) The twelve hours in the clock face are to be colored. In how many ways can this be done if
d) 6 of them must be white, 4 must be black and 2 must be red?
Answer: we see three tasks, clearly. The answer is (12C6) (6C4) (2C2)=13860
e) 6 of them must be white, the others may be blue or red?
Answer: we see two tasks, clearly. The answer is (12C6) . 26 = 59136
f) the even hours are white, the odd ones alternate between blue and red?
Answer: There is just one way (!) to paint the even hours (and not 12C6 ). Due to the alternating
requirement, there are two ways to complete the coloring: BRBRBR and RBRBRB for 1-3-5-7-9-11. So
the final answer is 2.
8) Prove that the product of n consecutive natural numbers is always a multiple of n!
Proof:
the
product
of
n
consecutive
numbers
can
be
written
as
(r+1).(r+2). . . (r+n) and we need to show that there is an integer x such that
(n!) x = (r+1).(r+2). . . (r+n). From this equation we infer that such an integer x must satisfy x= (r+1).(r+2). .
. (r+n) /n! Multiply both numerator and denominator by r!, obtaining x = r! (r+1).(r+2). . . (r+n)/r!n!
Take a close look at the numerator: it coincides with (r+n)! Therefore x = r+nCr is the solution to our
equation. Since r+nCr is an integer (being a combinatorial) x is an integer, as sought.
Remark: I invite you to illustrate this proposition by considering some numerical examples. Choose the n
you most like, and then the n consecutive naturals of your choice. Use your calculator to verify that n!
divides their product.
9) Eight people are standing on the ground floor of a building, waiting for the elevator. If the elevator can
carry up to 5 people, in how many ways can it leave the ground floor? (Note: the order in which people
enter the elevator does not matter)
Answer: observe that the different ways correspond to the number of different subsets of eight individuals,
up to a size of 5. Thus, we have 8C0 + 8C1 + 8C2 + 8C3 + 8C4 + 8C5 = 1+8+28+56+70+56=219.
Comment: we are allowing the elevator to leave empty: that is the reason for counting 8C0
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10) Problems involving teams. Notice that the ordering is irrelevant (unless otherwise specified. Hence
{a,b} as a team is equal to {b,a}, where a and b are persons.
All these problems refer to a group G of twelve people, that is, G is the “pool”.
I)
a) How many five-person teams can be chosen from G?
Answer: 12C5 = 792
b) Suppose two members (A and B) of G insist on working as a pair. That is, the teams must contain both
of them or neither of them? How many five-person teams can now be chosen?
Answer: the number of teams that contain both A and B is 10C3 = 120.
The number of teams that contain neither A nor B is 10C5 = 252. The total is 120 + 252 = 372.
c) On the contrary, supposed that A and B do not get along and so refuse to work together on a team. How
many five-person teams can now be formed?
Answer: the number is 792 – 120 = 672, that is, all the teams (see a) minus those in b) that contain both A
and B . Do you see why?
II)Let us now be more specific: we know that G contains five men and seven women. Use this additional
information to answer these questions.
d) How many five-person teams can be chosen consisting of three men and two women?
Answer: 5C3 × 7C2 = 210
e) How many five-person teams consist of at least one man?
Answer: We compute the number of five-person teams (by a it is 792) and subtract the number of teams
with no men, that is, with only women as members: this is 7C5 =21. So the final answer to the problem is
771.
f) How many five-person teams contain at most one man?
Recall: at least one, means 1 or more. At most one means 0 or 1.
Answer: The number of five-person teams with exactly one man is 5 × 7C4 = 175
As seen above, the number of teams with no men, that is, with only women as members is 7C5 =21.
So the final answer is 175 + 21 = 196
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Dr. J. Viola-Prioli