How many different possible outcomes are there for a swimming event with six competitors?
\(6! = \text{720}\)
10.6 Application to counting problems
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10.5 Factorial notation
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10.7 Application to probability problems
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10.6 Application to counting problems (EMCK4)
Worked example 14: Further arrangement of outcomes without repetition
Eight athletes take part in a \(\text{400}\) \(\text{m}\) race. In how many different ways can the first three places be arranged?
Eight different athletes can occupy the first \(\text{3}\) places. For the first place, there are \(\text{8}\) different choices. For the second place there are \(\text{7}\) different choices and for the third place there are \(\text{6}\) different choices. Therefore \(\text{8}\) different athletes can occupy the first three places in: \[8 \times 7 \times 6 = 336 \text{ ways}\]
Worked example 15: Arrangement of objects with constraints
In how many ways can seven boys of different ages be seated on a bench if:
- the youngest boy sits next to the oldest boy?
- the youngest and the oldest boys must not sit next to each other?
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This question is a little different to the previous problems of arrangements without repetition. In this question, we have the constraint that the youngest boy and the oldest boy must sit together. The easiest way to think about this, is to see each set of objects which have to be together as a single object to arrange. \(\\\)
If we let boy \(= B\) and let the number subscript indicate order of age, we can view the objects to arrange as follows:
\[\begin{array}{cccccc} (B_{1};B_{7}); & (B_{2}); & (B_{3}); & (B_{4}); & (B_{5}); & (B_{6}) \\ 1 & 2 & 3 & 4 & 5 & 6 \end{array}\]If the the youngest and oldest boys are treated as a single object, there are six different objects to arrange so there are \(6!\) different arrangements. However, the youngest and oldest boys can be arranged in \(2!\) different ways and still be together:
\[\begin{array}{ccc} (B_{1};B_{7}) & \text{or} & (B_{7};B_{1}) \end{array}\]Therefore there are:
\[6! \times 2! = \text{1 440} \text{ ways for the boys to be seated}\] -
The arrangements where the youngest and oldest must not sit together is the total number of arrangements minus the number of arrangements where the oldest and youngest sit together. Therefore, there are:
\[7! - 1440 = \text{3 600} \text{ ways for the boys to be seated}\]
Number of choices in a row
Textbook Exercise 10.6
How many different possible outcomes are there for the gold (1st), silver (2nd) and bronze (3rd) medals in a swimming event with six competitors?
\(6 \times 5 \times 4 = \text{120}\)
Susan wants to visit her friends in Pretoria, Johannesburg, Phalaborwa, East London and Port Elizabeth. In how many different ways can the visits be arranged?
\(5! = \text{120} \text{ ways}\)
A head boy, a deputy head boy, a head girl and a deputy head girl must be chosen out of a student council consisting of 18 girls and 18 boys. In how many ways can they be chosen?
\(18 \times 17 + 18 \times 17 = \text{612} \text{ ways}\)
Twenty different people enter a golf competition. Only the first six of them can win prizes. In how many different ways can the prizes be won?
\(20 \times 19 \times 18 \times 17 \times 16 \times 15 = \text{27 907 200} \text{ ways}\)
Three letters of the word 'EMPTY' are arranged in a row. How many different arrangements are possible?
\(5 \times 4 \times 3 = \text{60} \text{ arrangements}\)
Pool balls are numbered from \(\text{1}\) to \(\text{15}\). You have only one set of pool balls. In how many different ways can you arrange:
all \(\text{15}\) balls. Write your answer in scientific notation, rounding off to two
decimal places.
\(\text{15}! = \text{1,31} \times \text{10}^{\text{12}}\)
four of the \(\text{15}\) balls.
\(15 \times 14 \times 13 \times 12= \text{32 760}\)
The captains of all the sports teams in a school have to stand next to each other for a photograph. The school sports programme offers rugby, cricket, hockey, soccer, netball and tennis.
In how many different orders can they stand in the photograph?
\[6! = \text{720} \text{ different orders}\]
In how many different orders can they stand in the photograph if the rugby captain
stands on the extreme left and the cricket captain stands on the extreme right?
Since we have no choice about where to put the rugby and cricket captains, there are only \(\text{4}\) people left to arrange.
\[4! = \text{24} \text{ different orders}\]In how many different orders can they stand if the rugby captain, netball captain and
cricket captain must stand next to each other?
The rugby captain, netball captain and cricket captain are treated as a single object, as they must stand together. So there are four different objects to arrange therefore there are \(4!\) different arrangements. The rugby captain, netball captain and cricket captain can also swop positions between themselves in \(3!\) different ways, therefore there are:
\[4! \times 3! = \text{144} \text{ different orders}\]How many three-digit numbers can be made from the digits 1 to 6 if:
repetition is not allowed?
\[6 \times 5 \times 4 = \text{120}\]
repetition is allowed?
\[6^{3}= \text{216}\]
There are two different red books and three different blue books on a shelf.
In how many different ways can these books be arranged?
\[5! = \text{120} \text{ different ways to arrange the books}\]
If you want the red books to be together, in how many different ways can the books be
arranged?
If the red books are treated as a single object, there are four different objects to arrange therefore there are \(4!\) different arrangements. The red books can also be rearranged between themselves in \(2!\) different ways, therefore there are:
\[4! \times 2! = \text{48} \text{ different ways to arrange the books}\]If you want all the red books to be together and all the blue books to be together, in
how many different ways can the books be arranged?
There are two groups of books, red and blue, which can be arranged \(2!\) ways. Then there are two red books which can be arranged in \(2!\) ways and there are three blue books which can be arranged in \(3!\) ways. Therefore, there are
\[2! \times 2! \times 3! = \text{24} \text{ different ways to arrange the books}\]There are two different Mathematics books, three different Natural Sciences books, two different Life Sciences books and four different Accounting books on a shelf. In how many different ways can they be arranged if:
the order does not matter?
\[11! = \text{39 916 800} \text{ ways to arrange the books}\]
all the books of the same subject stand together?
There are four groups of books, which can be arranged in \(4!\) different ways. Of the books, two are
Mathematics books, three are Natural Sciences books, two are Life Sciences books and four are Accounting
books. Therefore, there are:
\[4! \times 2! \times 3! \times 2! \times 4! = \text{13 824} \text{ ways to arrange the books}\]
the two Mathematics books stand first?
The Mathematics books can be arranged in \(2!\) ways while the remaining books can be arranged in \(9!\) ways. Therefore, there are:
\[2! \times 9! = \text{725 760} \text{ ways to arrange the books}\]the Accounting books stand next to each other?
The Accounting books can be arranged in \(4!\) ways and, if treated as a single object, can be arranged with the remaining books in \(8!\) ways. Therefore, there are:
\[4! \times 8! = \text{967 680}\]Worked example 16: Arrangement of letters
If you take the word, 'OMO', how many letter arrangements can we make if:
- we consider the two O’s as different letters?
- we consider the two O’s as identical letters?
-
Since we consider the two O’s as different letters, write the first O as \(\text{O}_{1}\) and the second \(\text{O}_{2}\). The different letter arrangements are as follows:
\[\begin{array}{ccc} \text{O}_{1} \text{M}\text{O}_{2}& \text{M}\text{O}_{1}\text{O}_{2} & \text{O}_{1} \text{O}_{2}\text{M} \\ \text{O}_{2} \text{M}\text{O}_{1}& \text{M}\text{O}_{2}\text{O}_{1} & \text{O}_{2} \text{O}_{1}\text{M} \end{array}\]We can see from writing out all the arrangements that there are \(\text{6}\) different ways for the letters to be arranged. This is not practical if there are a large number of letters. Instead, we can work this out more easily using the fundamental counting principle.
Using the fundamental counting principle, as there are \(\text{3}\) letters in the word OMO if we treat each O as a separate letter, there are \(3! = 6\) different arrangements.
-
If we consider the two O’s as identical letters, only \(\text{3}\) arrangements are possible:
\[\begin{array}{ccc} \text{OMO} & \text{MOO} & \text{OOM} \end{array}\]We can also work this out using a modified version of the previous identity. We know that if we treat each letter as different, the number of arrangements is \(3!\). However, when we have duplicate letters, we have to remove the identical arrangements of these letter from our final answer. So we divide by the factorial of the number of times a letter is repeated.
In this example, O appears twice so we divide \(3!\) by \(2!\): \[\text{number of arrangements} = \frac{3!}{2!} = 3\]
Worked example 17: The number of letter arrangements for a longer word
If you take the word 'BASSOON', how many letter arrangements can you make if:
- repeated letters are treated as different?
- repeated letters are treated as identical?
- the word starts with an O and repeated letters are treated as identical?
- the word starts and ends with the same letter and repeated letters are treated as identical?
-
There are \(\text{7}\) letters in the word 'BASSOON'. If we treat each letter as a different letter, there are \(7! = \text{5 040}\) arrangements.
- If repeated letters are treated as identical characters, there are two S's and two O's. This is similar to the previous worked example except now we have more than one letter repeated. When more than one letter is repeated, we have to divide the total number of possible arrangements by the product of the factorials of the number of times each letter is repeated.\[\text{Number of arrangements } = \frac{7!}{2! \times 2!} = \text{1 260} \text{ arrangements}\]
- If the word starts with an 'O', there are still 6 letters left of which two are S's. \[\text{Number of arrangements } = \frac{6!}{2!} = 360 \text{ arrangements}\]
-
If the word starts and ends with the same letter, there are two possibilities:
- S - - - - - S with the letters in between consisting of 'B','A','O','O' and 'N'. Therefore: \[\text{Number of arrangements } = \frac{5!}{2!} = 60 \text{ arrangements}\]
- O - - - - - O with the letters in between consisting of 'B','A','S','S' and 'N'. Therefore: \[\text{Number of arrangements } = \frac{5!}{2!} = 60 \text{ arrangements}\]
Therefore, the total number of arrangements \(= 60 + 60 = 120\).
This allows us to formulate the following:
For a set of \(n\) objects, of which \(n_{1}\) are the same, \(n_{2}\) are the same \(\ldots, n_{k}\) are the same, the number of arrangements \(= \dfrac{n!}{n_{1}! \times n_{2}! \dots n_{k}!}\)
Number of arrangements of sets containing alike objects
Textbook Exercise 10.7
You have the word 'EXCELLENT'.
If the repeated letters are regarded as different letters, how many letter arrangements
are possible?
\[9! = \text{362 880}\]
If the repeated letters are regarded as identical, how many letter arrangements are
possible?
There are 3 E's and 2 L's so we have to divide by \(3!\) and \(2!\).
\[\dfrac{9!}{3! \times 2!} = \text{30 240}\]If the first and last letters are identical, how many letter arrangements are there?
The word could start and end in E or L. With an E, there are \(\dfrac{7!}{2!}\) letter arrangements
(divide by 2 L's) and with an L there are \(\dfrac{7!}{3!}\) letter arrangements (divide by 3 E's).
Therefore, there are:
\[\dfrac{7!}{3!} + \dfrac{7!}{2!} = \text{840} + \text{2 520} = \text{3 360} \text{ possible
letter arrangements}\]
How many letter arrangements can be made if the arrangement starts with an L?
This is equivalent to removing one L from the letters available for arrangement. Therefore, there are:
\[\dfrac{8!}{3!} = \text{6 720} \text{ possible letter arrangements}\]
How many letter arrangements are possible if the word ends in a T?
This is equivalent to removing the T from the letters available for arrangement. Therefore, there are:
\[\dfrac{8!}{3! \times 2!} = \text{3 360} \text{ possible letter arrangements}\]
You have the word 'ASSESSMENT'.
If the repeated letters are regarded as different letters, how many letter arrangements
are possible?
\[10! = \text{3 628 800}\]
If the repeated letters are regarded as identical, how many letter arrangements are
possible?
\[\dfrac{10!}{4! \times 2!} = \text{75 600}\]
If the first and last letters are identical, how many letter arrangements are there?
The word could start and end in S or E. With an S, there are \(\dfrac{8!}{2! \times 2!}\) letter
arrangements (divide by 2 E's and 2 remaining S's) and with an E there are \(\dfrac{8!}{4!}\) letter
arrangements (divide by 4 S's). Therefore, there are:
\[\dfrac{8!}{2! \times 2!} + \dfrac{8!}{4!} = \text{10 080} ~ + \text{1 680} =
\text{11 760} \text{ possible letter arrangements}\]
How many letter arrangements can be made if the arrangement starts with a vowel?
The word could start with A or E. With an A, there are \(\dfrac{9!}{4! \times 2!}\) letter arrangements
(divide by 2 E's and 4 S's) and with an E there are \(\dfrac{9!}{4!}\) letter arrangements (divide by 4
S's). Therefore, there are:
\[\dfrac{9!}{4! \times 2!} + \dfrac{9!}{4!} = \text{7 560} ~ + \text{15 120} =
\text{22 680} \text{ possible letter arrangements}\]
How many letter arrangements are possible if all the S's are at the beginning of the
word?
This is equivalent to removing all the S's from the letters available for arrangement. Therefore, there
are:
\[\dfrac{6!}{2!} = \text{360} \text{ possible letter arrangements}\]
On a piano the white keys represent the following notes: C, D, E, F, G, A, B. How many tunes, seven notes in length, can be composed with these notes if:
a note can be played only once?
\[7! = \text{5 040} \text{ possible tunes}\]
the notes can be repeated?
\[7^{7} = \text{823 543} \text{ possible tunes}\]
the notes can be repeated and the tune begins and ends with a D?
The tune starting and ending with a D leaves five possible positions in which to arrange the seven notes.
Therefore, there are:
\[7^{5} = \text{16 807} \text{ possible tunes}\]
the tune consists of 3 D’s, 2 B’s and 2 A’s.
\[\dfrac{7!}{3! \times 2! \times 2!} = \text{210} \text{ possible tunes}\]
There are three black beads and four white beads in a row. In how many ways can the beads be arranged if:
same-coloured beads are treated as different beads?
\[7! = \text{5 040} \text{ ways}\]
same-coloured beads are treated as identical beads?
\[\dfrac{7!}{3! \times 4!} = \text{35} \text{ ways}\]
There are eight balls on a table. Some are white and some are red. The white balls are all identical and the red balls are all identical. The balls are removed one at a time. In how many different orders can the balls be removed if:
seven of the balls are red?
\[\dfrac{8!}{7!} = \text{8} \text{ different orders}\]
three of the balls are red?
\[\dfrac{8!}{3! \times 5!} = \text{56} \text{ different orders}\]
there are four of each colour?
\[\dfrac{8!}{4! \times 4!} = \text{70} \text{ different orders}\]
How many four-digit numbers can be formed with the digits 3, 4, 6 and 7 if:
there can be repetition?
\[4^{4} = \text{256} \text{ possible numbers}\]
each digit can only be used once?
\[4! = \text{24} \text{ possible numbers}\]
if the number is odd and repetition is allowed?
For the number to be odd, it must end in 3 or 7. For the numbers ending in 3, there are \(4^{3}\)
different arrangements of the first three digits and similarly, for 7, there are \(4^{3}\) different
arrangements. Therefore, there are
\[4^{3} + 4^{3} = \text{128} \text{ possible numbers which match the criteria}\]
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10.5 Factorial notation
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10.7 Application to probability problems
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