Notes on permutations and combinations
Download as DOC, PDF1 like6,861 views
The document defines permutations and combinations. Permutations refer to arrangements where order matters, while combinations refer to selections where order does not matter. Formulas for calculating permutations and combinations are provided. Several examples illustrate calculating the number of possible arrangements and selections in different scenarios involving people, objects, and committees.
![then S, or T then S then H, or H-T-S or H-S-T. Note that we have 3 choices for the first
selection, 2 for the second, and 1 for the third. So the number of different ways of selecting the
objects is 3∙2∙1 or 3!. We know from example 5 that if we choose S first, there are 6 possible
arrangements of the distinguishable objects when order of selection doesn’t matter. If we
multiply that by the number of ways the objects can be ordered, then we have 6∙3! = 36 different
arrangements of the objects when order of selection matters.
Example P7:
Now suppose that you have 3 squares, 4 triangles, and 7 hexagons, and each of the 14 objects is
a different color. We learned in Ex. 1 that the arrangement of n distinguishable objects is n!.
Suppose also that the squares must be grouped together, the triangles must be grouped together,
and the hexagons must be grouped together. In how many different ways can each type of object
be arranged? The group of 3 distinguishable squares can be arranged in 3! ways, the 4
distinguishable triangles in 4! ways, and the 7 distinguishable hexagons in 7! ways. Given that
we have 3 groups of distinguishable objects, we can arrange those groups in 3! ways. Thus, the
number of distinguishable arrangements of the objects when they must be grouped by type is
3! ∙ 4! ∙ 7! ∙ 3! = 4,354,560
Definition:
A combination of n objects taken r at a time is a selection of r objects from a total of n objects
(r ≤ n), where ORDER DOES NOT MATTER. The notation is usually
n
r
and read, “n choose r.”
Formula: The general formula for combinations of n choose r
C(n, r) = n!/[(n – r)!∙r!].
Example C1—r = n: How many ways can a committee of 7 people be drawn from a group of 7
people?
C(7, 7) = 7!/[(7 – 7)!∙7!] = 7!/[0!∙7!] = 7!/7! = 1.
In other words, there is exactly 1 way to choose all 7 people to be on the committee since order
doesn’t matter.
Example C2
—r < n: How many ways can 2 co-chairpersons be chosen from a committee of 7 people?
C(7, 2) = 7!/[(7 – 2)!∙2!] = 7!/[5!∙2!] = 7∙6∙5∙4∙3∙2∙1/(5∙4∙3∙2∙1 )(2∙1) = 7∙6/2∙1 = 7∙3 = 21.
In other words, there are 21 ways to select a pair of co-chairs from a committee of 7 people.
Relating the above scenario to permutations, note that there are 7 ways to choose the first co-
chair and 6 ways to choose the second co-chair. So if order mattered, we would have 42
possible choices. However, the combination of 2 co-chairs is the same, no matter which person
was selected first. Thus, we must divide by the by the 2! ways the two chair persons could have
been selected. Applying the actual formula, we have
P(7, 2) = 7!/(7 – 2)! = 7!/5! = 7∙6 = 42.](https://image.slidesharecdn.com/notesonpermutationsandcombinations-120702071018-phpapp01/85/Notes-on-permutations-and-combinations-3-320.jpg)
![But
C(7, 2) = 7!/[(7 – 2)!∙2!] = 7!/[5! ∙2!] = (7∙6)/(2∙1) = 42/2 = 21.
Example C3:
In a club with 8 males and 11 female members, how many 5-member committees can be chosen
that have
(a) 4 females?
Solution: Since order doesn’t matter, we will be using our combinations formula.
How many ways can the 4 females be chosen from the 11 females in the club?
C(11, 4) = 11!/[(11 – 4)! ∙4!] = 11!/(7! ∙4!) = (11∙10∙9∙8)/ (4∙3∙2∙1) = 330 ways.
But this gives us only the 4 female members for our 5-member committee. For a 5-member
committee with 4 females, how many males must be chosen? Only 1. How many ways can
the 1 male be chosen from the 8 males in the club?
C(8, 1) = 8!/[(8 – 1)! ∙1!] = 8!/(7! ∙1!) = 8 ways.
Thus we have 8∙330 = 2640 ways to select a 5-member committee with exactly 4
female members from this club.
(b) At least 4 females?
Solution: In part (a) we found that there are 2640 ways to select exactly 4 females for our 5-
member committee. If we need at least 4 females, we could have 4 females or 5 females,
couldn’t we? How many ways can we select 5 females from the 11 females in the club,
given that order doesn’t matter?
C(11, 5) = 11!/[(11 – 5)! ∙5!] = 11!/(6! ∙5!) = (11∙10∙9∙8∙7)/ (5∙4∙3∙2∙1) = 462 ways.
So to have at least 4 females, we could have C(11, 4) ∙C(8, 1) or C(11, 5). Since there are
330 ways to achieve the first and 462 ways to achieve the second, we have
330 + 462 = 792 ways
to select a 5-member committee with at least 4 female members from the club.
(c) No more than 2 males?
Solution: “No more than 2” males means we could have 1 male or 2 males on our 5-member
committee. Since we have the condition of the 5-member committee, we have the following
possibilities: (1) 1 male and 4 females, or (2) 2 males and 3 females. Since order doesn’t
matter, we have C(8, 1)∙C(11, 4) ∪ C(8, 2)∙C(11, 3) ways to select the committee. Applying
the combinations formulas, we have
{8!/[(8 – 1)! ∙1!]} ∙ {11!/[(11 – 4)! ∙4!]} + {8!/[(8 – 2)! ∙2!]} ∙ {11!/[(11 – 3)! ∙3!]} =
(8)(330) + (28)(165) = 2540 + 4620 = 7260 ways
to select a 5-member committee with no more than 2 male members from the club.
Example C4:
Suppose we have the same club with 8 males and 11 females, but this time we want to know
how many committees could be chosen with at least 1 but no more than 3 members, regardless of
gender. Therefore, committees must have 1, 2, or 3 members. What are the possible committee
memberships?
Solution: The possibilities are 1M or 1F or 2M or 2F or 1M & 1F or 1M & 2F or 2M & 1F or
3M or 3F. Does order matter? No, it does not. So we use our combinations formulas again for
each possible committee make-up and add them to get the solution to our problem.](https://image.slidesharecdn.com/notesonpermutationsandcombinations-120702071018-phpapp01/85/Notes-on-permutations-and-combinations-4-320.jpg)
Notes on permutations and combinations
- 1. PERMUTATIONS AND COMBINATIONS Definition: A permutation of n objects taken r at a time is a selection of r objects from a total of n objects (r ≤ n), where ORDER MATTERS. Example: P1-- r = n: Suppose you have 6 (distinguishable) people to seat in 6 chairs. The first person may choose her/his chair from 6 possibilities. The second person must choose from the 5 remaining chairs, etc., until the 6th person has to take the only chair that’s left. This arrangement can be done in 6∙5∙4∙3∙2∙1, or 6! (“six factorial”) ways. (See #23.a. on p. 401. Although this problem is somewhat ambiguous, assume that each of the 14 objects, not each type of object, is a different color and therefore distinguishable from the others.) Example P2— r < n: Now suppose that you have 6 (distinguishable) people but only 4 chairs; so only 4 of the 6 people will be able to sit down. How many arrangements of the 6 people can be seated in the 4 chairs? This arrangement can be done in 6∙5∙4∙3 ways. Formulas: The general formula for a permutation of n objects arranged r at a time is P(n, r) = n!/(n - r)! Note that in Ex. P1, r = n, so n!/(n - r)! = n!/(n - n)! = n!/0! Since 0! is defined to be 1, we have n!/1 = n!. Therefore, the answer to P1 is simply 6!. Applying the formula to Ex. P2 yields 6!/(6 - 4)! = 6!/2! = (6∙5∙4∙3∙2∙1)/(2∙1) = 6∙5∙4∙3. Thus, the general formula applies in both cases, whether r = n or r < n. Example P3: Suppose that you have 3 objects, but they are not all distinguishable. For example, suppose that you have 1 red square and 2 green triangles. How many different arrangements of those 3 objects are possible? If the two green triangles switched places with each other, would the arrangements be distinguishable from the ones given above? No, they would not. Therefore, the 2! ways the green triangles could be rearranged must be eliminated. So if we have 3 objects, where 2 objects are the same, then the number of possible arrangements is P(3, 3) = 3!/2! = (3∙2∙1)/(2∙1) = 3 (not 3!).
- 2. Example P4: Suppose that you have 6 objects—1 red square, 2 green triangles, and 3 yellow hexagons. How many different arrangements of the 6 objects are possible? In this case, you have 2 indistinguishable green triangles and 3 indistinguishable yellow hexagons. Note that you have 2! ways to interchange the 2 triangles and 3! ways to interchange the 3 hexagons. So you must take those out of the total possible permutations. Thus, P(6, 6) = 6!/2!3! = (6∙5∙4∙3∙2∙1)/(2∙1)(3∙2∙1) = (6∙5∙4)/2 = 60 (not 6!). Example P5: Suppose you have the same set of 6 objects given in example P4 above, but this time each triangle is a different shade of green and each hexagon is a different shade of yellow so that they are distinguishable. This time you want to select exactly one object of each type. (Assume the different types are in 3 separate bags.) How many different ways can this be done if the order in which the objects are selected doesn’t matter? Here are the possibilities (i.e., the sample space): There is 1 way to get a red square, 2 ways to get a green triangle, and 3 ways to get a yellow hexagon; so the total number of arrangements (since the order of selection doesn’t matter) is 1∙2∙3 = 6. Example P6: Suppose that you have the same set of objects given in Ex. 5, but this time the order of selection makes a difference. How many ways can we order the selection? We can choose square (S) then triangle (T) then hexagon (H), or we can choose S then H then T, or T then H
- 3. then S, or T then S then H, or H-T-S or H-S-T. Note that we have 3 choices for the first selection, 2 for the second, and 1 for the third. So the number of different ways of selecting the objects is 3∙2∙1 or 3!. We know from example 5 that if we choose S first, there are 6 possible arrangements of the distinguishable objects when order of selection doesn’t matter. If we multiply that by the number of ways the objects can be ordered, then we have 6∙3! = 36 different arrangements of the objects when order of selection matters. Example P7: Now suppose that you have 3 squares, 4 triangles, and 7 hexagons, and each of the 14 objects is a different color. We learned in Ex. 1 that the arrangement of n distinguishable objects is n!. Suppose also that the squares must be grouped together, the triangles must be grouped together, and the hexagons must be grouped together. In how many different ways can each type of object be arranged? The group of 3 distinguishable squares can be arranged in 3! ways, the 4 distinguishable triangles in 4! ways, and the 7 distinguishable hexagons in 7! ways. Given that we have 3 groups of distinguishable objects, we can arrange those groups in 3! ways. Thus, the number of distinguishable arrangements of the objects when they must be grouped by type is 3! ∙ 4! ∙ 7! ∙ 3! = 4,354,560 Definition: A combination of n objects taken r at a time is a selection of r objects from a total of n objects (r ≤ n), where ORDER DOES NOT MATTER. The notation is usually n r and read, “n choose r.” Formula: The general formula for combinations of n choose r C(n, r) = n!/[(n – r)!∙r!]. Example C1—r = n: How many ways can a committee of 7 people be drawn from a group of 7 people? C(7, 7) = 7!/[(7 – 7)!∙7!] = 7!/[0!∙7!] = 7!/7! = 1. In other words, there is exactly 1 way to choose all 7 people to be on the committee since order doesn’t matter. Example C2 —r < n: How many ways can 2 co-chairpersons be chosen from a committee of 7 people? C(7, 2) = 7!/[(7 – 2)!∙2!] = 7!/[5!∙2!] = 7∙6∙5∙4∙3∙2∙1/(5∙4∙3∙2∙1 )(2∙1) = 7∙6/2∙1 = 7∙3 = 21. In other words, there are 21 ways to select a pair of co-chairs from a committee of 7 people. Relating the above scenario to permutations, note that there are 7 ways to choose the first co- chair and 6 ways to choose the second co-chair. So if order mattered, we would have 42 possible choices. However, the combination of 2 co-chairs is the same, no matter which person was selected first. Thus, we must divide by the by the 2! ways the two chair persons could have been selected. Applying the actual formula, we have P(7, 2) = 7!/(7 – 2)! = 7!/5! = 7∙6 = 42.
- 4. But C(7, 2) = 7!/[(7 – 2)!∙2!] = 7!/[5! ∙2!] = (7∙6)/(2∙1) = 42/2 = 21. Example C3: In a club with 8 males and 11 female members, how many 5-member committees can be chosen that have (a) 4 females? Solution: Since order doesn’t matter, we will be using our combinations formula. How many ways can the 4 females be chosen from the 11 females in the club? C(11, 4) = 11!/[(11 – 4)! ∙4!] = 11!/(7! ∙4!) = (11∙10∙9∙8)/ (4∙3∙2∙1) = 330 ways. But this gives us only the 4 female members for our 5-member committee. For a 5-member committee with 4 females, how many males must be chosen? Only 1. How many ways can the 1 male be chosen from the 8 males in the club? C(8, 1) = 8!/[(8 – 1)! ∙1!] = 8!/(7! ∙1!) = 8 ways. Thus we have 8∙330 = 2640 ways to select a 5-member committee with exactly 4 female members from this club. (b) At least 4 females? Solution: In part (a) we found that there are 2640 ways to select exactly 4 females for our 5- member committee. If we need at least 4 females, we could have 4 females or 5 females, couldn’t we? How many ways can we select 5 females from the 11 females in the club, given that order doesn’t matter? C(11, 5) = 11!/[(11 – 5)! ∙5!] = 11!/(6! ∙5!) = (11∙10∙9∙8∙7)/ (5∙4∙3∙2∙1) = 462 ways. So to have at least 4 females, we could have C(11, 4) ∙C(8, 1) or C(11, 5). Since there are 330 ways to achieve the first and 462 ways to achieve the second, we have 330 + 462 = 792 ways to select a 5-member committee with at least 4 female members from the club. (c) No more than 2 males? Solution: “No more than 2” males means we could have 1 male or 2 males on our 5-member committee. Since we have the condition of the 5-member committee, we have the following possibilities: (1) 1 male and 4 females, or (2) 2 males and 3 females. Since order doesn’t matter, we have C(8, 1)∙C(11, 4) ∪ C(8, 2)∙C(11, 3) ways to select the committee. Applying the combinations formulas, we have {8!/[(8 – 1)! ∙1!]} ∙ {11!/[(11 – 4)! ∙4!]} + {8!/[(8 – 2)! ∙2!]} ∙ {11!/[(11 – 3)! ∙3!]} = (8)(330) + (28)(165) = 2540 + 4620 = 7260 ways to select a 5-member committee with no more than 2 male members from the club. Example C4: Suppose we have the same club with 8 males and 11 females, but this time we want to know how many committees could be chosen with at least 1 but no more than 3 members, regardless of gender. Therefore, committees must have 1, 2, or 3 members. What are the possible committee memberships? Solution: The possibilities are 1M or 1F or 2M or 2F or 1M & 1F or 1M & 2F or 2M & 1F or 3M or 3F. Does order matter? No, it does not. So we use our combinations formulas again for each possible committee make-up and add them to get the solution to our problem.
- 5. C(8, 1) + C(11, 1) + C(8, 2) + C(11, 2) + C(8, 1)∙C(11, 1) + C(8, 1) ∙C(11, 2) + C(8, 2) ∙C(11, 1) + C(8, 3) + C(11, 3). we can calculate the answer using the nCr command on your calculator or the combin(n, r) using MathCAD.