Factorial notation

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  1. How many ways are there to arrange 4 different books side by side on a bookshelf?
    1. factorial notation: n! = n (n-1) (n-2) (n-3) (n-4) . . . . . (5) (4) (3) (2) (1)
      by definition : 0! = 1
      1. Evaluate: 5!
      2. Evaluate: 7!5!\frac{{7!}}{{5!}}
      3. Simplify: (n+3)!n!\frac{{\left( {n + 3} \right)!}}{{n!}}
      4. Simplify: (n−1)!(n+2)!\frac{{\left( {n - 1} \right)!}}{{\left( {n + 2} \right)!}}
      5. Simplify: (n+1)!    (n−3)!(n!)2\frac{{\left( {n + 1} \right)!\;\;\left( {n - 3} \right)!}}{{{{(n!)}^2}}}
      6. Solve: n!(n−2)!    3!=7\frac{{n!}}{{\left( {n - 2} \right)!\;\;3!}} = 7
    2. arrangement of words "without repititions" = n!
      Determine the number of different arrangements of all the letters in the following words:
      1. DOG
      2. MATH
      3. COMPUTER
    3. arrangement of words "with repititions" = n!(1st  repetition)!      (2nd  repetition)!      (3rd  repetition)!  …..\frac{{n!}}{{\left( {{1^{st}}\;repetition} \right)!\;\;\;\left( {{2^{nd}}\;repetition} \right)!\;\;\;\left( {{3^{rd}}\;repetition} \right)!\; \ldots ..}}
      Determine the number of different arrangements of all the letters in the following words:
      1. ABC vs ABB
      2. CANADA
      3. BANANA
      4. REPETITION
    4. arrangement with restrictions: must deal with the restrictions first!
      1. Determine the number of different arrangements of all the letters in the word: COMPUTER
        (i) if there are no restrictions
        (ii) if the vowels must be together
        (iii) the vowels must not be together
      2. i) How many ways are there to arrange 3 Math books (Math 10, Math 11, Math 12), 2 physics books (Phys 11, Phys 12), and 5 English (Eng 8, Eng 9, Eng 10, Eng 11, Eng 12) on a bookshelf?
        ii) What if the books on each subject must be kept together?
    5. seating arrangement
      1. How many ways can 4 girls and 4 boys sit in a row, if:
        i) they can sit anywhere?
        ii) all the girls must sit together, and all the boys must sit together?
        iii) all the girls must sit together, while the boys can pick their own seats?
        iv) girls and boys alternate?
      2. There are 3 couples, and they need to sit together. How many different ways can these 3 couples sit in a row?
      3. There are 7 people A, B, C, D, E, F, and G sitting in a row. How many different seating arrangements are there, if:
        i) A must be to the left of B, but they do not need to sit together?
        ii) A and B must sit together?
        iii) A and B cannot sit together?
    6. Seating arrangements are considered to be different only when the positions of the people are different relative to each other.
      How many seating arrangements are possible for 7 people sitting around a round table?
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      Topic Notes
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      Factorial notation


      What is a factorial?


      A factorial is the multiplication of a number with all of those positive numbers below it. In other words, when you want to compute the factorial of a number, you multiply this number times all of the numbers smaller than it up to 1.

      This up to 1 is very important since the factorial of a negative number does not exist, and the factorial of 0 is defined as 1. Therefore, the numbers in a factorial are all positive integers. Consequently, the formal factorial definition says that a factorial is the product of a positive integer with all the consecutive positive integers that have a lower value than itself.
      Mathematically speaking, the factorial of a number is defined as:


      n!=(n)(n−1)(n−2)(n−3)...(2)(1)n! =(n)(n-1)(n-2)(n-3)... (2)(1)
      Equation 1: Factorial formula

      And so, the factorial notation of a number is done just by applying an exclamation mark to its right. Something important to remember is that since the 0 factorial is defined as 1 (0! = 1), then it means that the zero factorial is equal to the factorial of 1.

      Combinatorics problems using factorials


      On the next examples we will get to see how is that the factorial notation of a multiplication is useful when working in combinatorics problems.

      Example 1

      How many ways are there to arrange 4 different books side by side on a bookshelf?

      For this case we have four different books which we will define as a math book, a physics book, a chemistry book and a history book. And so, the setting is that we have four books that can be arranged in four different positions side by side as seen in the next figure:

      Factorial notation
      Figure 1: 4 books for 4 positions

      If you think about it, the first position has 4 different books as a possibility to be there, and we can pick any of them. And so, we pick one book from the four to be in the first position, and continue to the next position. For the second position, we are left with 3 possible books, we pick one and continue to the next position; on the third position we have 2 possible books left, we pick one and continue; in the last position (4th position) we have only one book left that can be there, and so our arrangement and the process we just explained can be seen in the next figure:

      Factorial notation
      Figure 2: Ways to arrange 4 different books side by side

      Thus, following the fundamental counting principle we know that we have to multiply the possible outcomes of each category to obtain a total of 24 possible ways in which the books can be arranged.

      If you notice, the multiplication resulted from the application of the fundamental counting principle is the same as the factorial of four! And so, we can clearly see how the factorial notation has been designed to help us out in abbreviating what can get to be a large multiplication into a simple expression such as: 4!

      Example 2

      Using what we have learned of factorial notation:
      n!=n(n−1)(n−2)(n−3)(n−4).....(5)(4)(3)(2)(1)n! = n (n-1) (n-2) (n-3) (n-4) . . . . . (5) (4) (3) (2) (1)
      And 0!=10! = 1 (zero factorial) by definition.

      a) \quad Evaluate: 5!
      This problem is very simple, you just have to compute the multiplication required to obtain the factorial of 5, using the factorial formula depicted in equation 1 we have that:

      5!=5×4×3×2×1=120 5! = 5 \times 4 \times 3 \times 2 \times 1 =120
      Equation 2: 5 Factorial


      b) \quad Evaluate: 7!5!\large \frac{7!}{5!}
      On this case you will see that when dividing factorials, if the factorial in the numerator is bigger than the one in the denominator, several of the factors above will be cancelled with those below and you can simplify the operation by computing the multiplication of the factors left:

      7!5!=7×6×5×4×3×2×15×4×3×2×1=\large \frac{7!}{5!} = \frac{7\times6\times5\times4\times3\times2\times1}{5\times4\times3\times2\times1} = 7×6=42 7\times6 = 42
      Equation 3: Dividing 7 factorial by 5 factorial


      c) \quad Simplify: (n + 3)!n!\large \frac{(n \, + \, 3)!}{n!}
      Another example of a factorial being divided by a smaller one. Notice how the factorial in the denominator cancels out most of the factors in the factorial above, and so, the expression gets simplified:

      (n + 3)!n!=(n + 3)(n + 2)(n + 1)(n)(n − 1)(n − 2)(n − 3)(n − 4)...(n)(n − 1)(n − 2)(n − 3)(n − 4)...= \large \frac{(n\,+\,3)!}{n!} = \frac{(n\,+\,3)(n\,+\,2)(n\,+\,1)(n)(n\,-\,1)(n\,-\,2)(n\,-\,3)(n\,-\,4)...}{(n)(n\,-\,1)(n\,-\,2)(n\,-\,3)(n\,-\,4)...} =

      (n+3)(n+2)(n+1) (n+3)(n+2)(n+1)

      (n+3)(n+2)(n+1)=(n2+2n+3n+6)(n+1)=(n2+5n+6)(n+1)(n+3)(n+2)(n+1)=(n^{2}+2n+3n+6)(n+1)=(n^{2}+5n+6)(n+1)

      (n2+5n+6)(n+1)=n3+n2+5n2+5n+6n+6=n3+6n2+11n+6(n^{2}+5n+6)(n+1)=n^{3}+n^{2}+5n^{2}+5n+6n+6=n^{3}+6n^{2}+11n+6

      therefore:

      (n + 3)!n!\large \frac{(n\,+\,3)!}{n!} =(n+3)(n+2)(n+1)=n3+6n2+11n+6= (n+3)(n+2)(n+1)=n^{3}+6n^{2}+11n+6
      Equation 4: Simplifying a division of factorials (higher factorial in the numerator)


      d) \quad Simplify: (n − 1)!(n + 2)!\large \frac{(n \, - \, 1)!}{(n\,+\,2)!}
      On this case we have two factorials dividing each other, but the highest one is found in the denominator, which means that it will cancel the factors contained in the numerator to produce an expression such as:

      (n − 1)!(n + 2)!=(n − 1)(n − 2)(n − 3)(n − 4)...(n+2)(n+1)(n)(n−1)(n−2)(n−3)(n−4)...=1(n+2)(n+1)(n)\large \frac{(n\,-\,1)!}{(n\,+\,2)!} = \frac{(n\,-\,1)(n\,-\,2)(n\,-\,3)(n\,-\,4)...}{(n+2)(n+1)(n)(n-1)(n-2)(n-3)(n-4)...} = \frac{1}{(n+2)(n+1)(n)}

      1(n+2)(n+1)(n)=1(n+2)(n2+n)=1n3+n2+2n2+n=1n3+3n2+n\large \frac{1}{(n+2)(n+1)(n)} = \frac{1}{(n+2)( n^{2}+n)} = \frac{1}{n^{3}+n^{2}+2n^{2}+n} = \frac{1}{n^{3}+3n^{2}+n}

      therefore:

      (n − 1)!(n + 2)!=1(n+2)(n+1)(n)=1n3+3n2+n \large \frac{(n\,-\,1)!}{(n\,+\,2)!} = \frac{1}{(n+2)(n+1)(n)} = \frac{1}{n^{3}+3n^{2}+n}
      Equation 5: Simplifying a division of factorials (higher factorial in the denominator)


      e) \quad Simplify: (n + 1)! (n − 3)!(n!)2\large \frac{(n \, + \, 1)! \, (n\,-\,3)!}{(n!)^{2}}
      For this case we combine what we have seen in parts c) and d) since this particular division of factorials includes multiplication of factorials in both the numerator and denominator. Therefore, you can factor out the whole expression into two divisions of factorials being multiplied.
      This particular operation can be done in many ways, let us show you one of them (you can practice at home by working the operations in different order).

      (n + 1)!(n − 3)!(n!)2=(n + 1)! × (n − 3)!n! × n!=(n + 1)!n!×(n − 3)!n!\large \frac{(n\,+\,1)! (n\,-\,3)!}{(n!)^{2}} = \frac{(n\,+\,1)! \, \times \, (n\,-\,3)!}{n! \, \times \, n!} = \frac{(n\,+\,1)!}{n!} \times \frac{(n\,-\,3)!}{n!}

      (n + 1)(n)(n − 1)(n − 2)(n − 3)...(n)(n − 1)(n − 2)(n − 3)... × (n − 3)(n − 4)(n − 5)...(n)(n − 1)(n − 2)(n − 3)(n − 4)(n − 5)...=n + 11 × 1(n)(n − 1)(n − 2)\large \frac{(n\,+\,1)(n)(n\,-\,1)(n\,-\,2)(n\,-\,3)...}{(n)(n\,-\,1)(n\,-\,2)(n\,-\,3)...} \, \times \, \frac{(n\,-\,3)(n\,-\,4)(n\,-\,5)...}{(n)(n\,-\,1)(n\,-\,2)(n\,-\,3)(n\,-\,4)(n\,-\,5)...} = \frac{n\,+\,1}{1} \, \times \, \frac{1}{(n)(n\,-\,1)(n\,-\,2)}

      n + 11 × 1(n)(n − 1)(n − 2)=n + 1(n)(n − 1)(n − 2)\large \frac{n\,+\,1}{1} \, \times \, \frac{1}{(n)(n\,-\,1)(n\,-\,2)} = \frac{n\,+\,1}{(n)(n\,-\,1)(n\,-\,2)}

      therefore:

      (n + 1)!(n − 3)!(n!)2=n + 1(n)(n − 1)(n − 2)\large \frac{(n\,+\,1)! (n\,-\,3)!}{(n!)^{2}} =\frac{n\,+\,1}{(n)(n\,-\,1)(n\,-\,2)}
      Equation 6: Simplifying a division of factorial multiplications


      f) \quad Solve: n!(n − 2)!3!\large \frac{n!}{(n\,-\,2)! 3!} =7=7
      In order to solve this particular factorial equation we pass the 3 factorial in the denominator to the other side of the equation by multiplying it in both sides, and so:

      n!(n − 2)!3!\large \frac{n!}{(n\,-\,2)! 3!} =7 = 7 \,   \, n!(n − 2)!\large \frac{n!}{(n\,-\,2)!} =7(3!)=7(3×2×1)=42= 7(3!)=7(3 \times 2 \times 1)=42
      Equation 7: Solving an equation with factorials (part 1)

      And now, we can just simplify the division of factorials on the left hand side in the same way as we have done it before:

      n!(n − 2)!=(n)(n − 1)(n − 2)(n − 3)(n − 4)...(n − 2)(n − 3)(n − 4)...=\large \frac{n!}{(n\,-\,2)!} = \frac{(n)(n\,-\,1)(n\,-\,2)(n\,-\,3)(n\,-\,4)...}{(n\,-\,2)(n\,-\,3)(n\,-\,4)...} = (n)(n − 1)=42(n)(n\,-\,1) = 42
      Equation 8: Solving an equation with factorials (part 2)

      Once the division of factorials has been simplified we obtain the next quadratic equation:

      (n)(n−1)=n2−n=42  (n)(n-1)=n^{2}-n=42 \;   n2−n−42=0 \; n^{2} -n-42=0
      Equation 9: Solving an equation with factorials (part 3)

      With this, we can use the quadratic formula and solve for nn. Since the quadratic formula is defined as:

      x= x = −b±b2−4ac2a\large \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}    \; when  ax2+bx+c=0 \; ax^{2} + bx + c = 0
      Equation 10: Quadratic formula

      On this case we have that  x=n, a=1, b=−1, c=−42\, x = n, \, a=1, \, b=-1, \, c = -42
      Therefore we solve for n in the following manner:

      n= n =−(−1)±(−1)2−4(1)(−42)2(1)=1±1+1682=1±1692=1±132 \large \frac{-(-1)\pm\sqrt{(-1)^{2}-4(1)(-42)}}{2(1)} = \frac{1\pm\sqrt{1+168}}{2} = \frac{1\pm\sqrt{169}}{2} = \frac{1\pm13}{2}

      And so, we have two options for nn:

      n=n = 142\large \frac{14}{2} =7  = 7 \; or  n= \; n = −122\large \frac{-12}{2} =−6 = - 6
      Equation 11: Solving an equation with factorials (part 4)

      Since the factorial of negative numbers does not exist, and the original expression to solve in this problem contains a term of n!n! in the numerator, nn must be equal to 7; putting nn = -6 would mean that we would have a (-6)! term in the numerator of the original expression, which is undefined! Therefore the answer is nn = 7

      Example 3

      In this example we take a look at the arrangement of letters in a word "without repetitions", which means to compute the n!n!, where nn equals the number of letters in the word.

      arrangement of words without repetitions = n!n!
      Equation 12: Different arrangements without repetition

      This process is the same that what we saw in the first example with the 4 different books being arranged side by side, you cannot repeat a book, just as in here the condition is that you do not repeat a letter. With all of this in mind, determine the number of different arrangements of all the letters in the following words:

      a) \quad DOG
      Total amount of letters = 3, therefore nn = 3, and the total amount of different positions in which these letters can be arranged is equal to 3! = (3)(2)(1) = 6

      b) \quad MATH
      Total amount of letters = 4, therefore nn = 4, and the total amount of different positions in which these letters can be arranged is equal to 4! = (4)(3)(2)(1) = 24

      c) \quad COMPUTER
      Total amount of letters = 8, therefore nn = 8, and the total amount of different positions in which these letters can be arranged is equal to 8! = (8)(7)(6)(5)(4)(3)(2)(1) = 40,320

      These examples were simple because these are words which have all distinct letters (without repetition), meaning that you can simply use the factorial of the amount of letters that you have in each case to obtain the different ways in which you can arrange them. The next example represents a case in which the words contain repeated letters, see how the difficulty level rises:

      Example 4

      On this example we take a look at the arrangement of letters in a word "with repetitions", which means that letters within the given word are not all distinct.
      The equation that produces the total amount of ways in which the letters of these words with repetitions can be arranged goes as:

      arrangement of words with repetitions  n!(1st repetition)! (2nd repetition)! (3rd repetition)!... \frac{arrangement\, of \,words \,with\, repetitions \; n!} {(1st\, repetition)! \, (2nd \,repetition)! \,(3rd\, repetition)! ... }
      Equation 13: Different arrangements with repetition

      Where nn equals the total number of letters in the word, and the repetitions mean how many times a letter repeats itself. For example if you have 5 letters and one repeats 3 times, then you would have that nn = 5 and the first repetition is 3! (there are no more repetitions in this case, and so, the equation for the arrangements would be simply: 5!/3!.

      With that in mind, determine the number of different arrangements of all the letters in the following words:

      a) \quad ABC vs ABB
      In this case, both ABC and ABB have nn = 3, therefore, for ABC we simply have that:

      arrangement of words without repetitions  =n!=3!=3 × 2 × 1=6 \, = n! = 3! =3 \, \times \, 2 \, \times \, 1 = 6
      Equation 14: Different arrangements without repetition

      For ABB we have that the letter B repeats twice, and so:

      arrangement of words with repetitions  3!2!=3 × 2 × 12 × 1= \, \large \frac{3!}{2!} = \frac{3\,\times\, 2\,\times\, 1}{2\,\times\,1} = 3 3
      Equation 15: Different arrangements with repetition


      b) \quad Canada
      The total arrangements possible with the letters in the word Canada are calculated as follows (where nn = 6):

      arrangement of words with repetitions

      =6!3!=6 × 5 × 4 × 3 × 2 × 13 × 2 × 1=\large = \frac{6!}{3!} = \frac{6\,\times\, 5 \,\times \, 4 \, \times \, 3 \, \times \, 2\, \times \, 1}{3 \, \times \, 2\, \times \, 1} = 6×5×4=1206 \times 5 \times 4 = 120
      Equation 16: Different arrangements of the letters in the word Canada


      c) \quad BANANA
      For this case nn =6 and you have two repetitions, letter N repeats twice, and letter A repeats trice; therefore:

      arrangement of words with repetitions

      =6!2! ⋅ 3!=6 × 5 × 4 × 3 × 2 × 1(2 × 1)(3 × 2 × 1)=\large = \frac{6!}{2! \, \cdot \, 3!} = \frac{6\,\times\, 5 \,\times \, 4 \, \times \, 3 \, \times \, 2\, \times \, 1}{(2\, \times \, 1)(3 \, \times \, 2\, \times \, 1)} = 6×5×2=606 \times 5 \times 2 = 60
      Equation 17: Different arrangements of the letters in the word banana


      d) \quad REPETITION
      For this case nn = 10 and you have three repetitions: the letters E, T and I repeat twice each, therefore:

      arrangement of words with repetitions =10!2!⋅2!⋅2!= \large \frac{10!}{2!\cdot 2!\cdot 2!}

      =10!2!⋅2!⋅2!=10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1(2 × 1)(2 × 1)(2 × 1)=\large = \frac{10!}{2!\cdot 2!\cdot 2!} = \frac{10\,\times\,9\,\times\,8\,\times\,7\,\times\,6\,\times\, 5 \,\times \, 4 \, \times \, 3 \, \times \, 2\, \times \, 1}{(2\, \times \, 1)(2 \, \times \, 1) (2 \, \times \, 1)} = 10×9×8×7×6×5×4×3=453,60010 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 = 453,600
      Equation 18: Different arrangements of the letters in the word banana


      Example 5

      For this problem example we will be taking a look at arrangement with restrictions, meaning that even when you have a word with all distinct letters, there may be certain conditions imposed on the arrangements of its words and so, these conditions (or restrictions) must be dealt with first.

      Determine the number of different arrangements of all the letters in the word: COMPUTER

      1. \quad if there are no restrictions
      In this case, we do not have repetitions and no restrictions, therefore, the arrangements are easy to calculate: just calculate factorial products for the total number of letters in the word. Since there are 8 letters in the word computer, then the total possible arrangements are 8! = 40,320.

      2. \quad if the vowels must be together
      We have 3 vowels (O, U and E) in the word COMPUTER, and our restriction for this case is that all of them will be written together. So how many arrangements of the word meet this restriction?

      Let us think on the word COMPUTER, take the three vowels out: O,U,E and think of them as a single unit to have: C, M, P, T, R, {O, U, E} as the six components of the word.
      Then, we use the same logic used in example one (about arranging books in different order side by side) and think of the letters in the word computer as a unit (just like each book in example one) to be arranged in different positions side by side, except for the vowels which are all merged together in a single unit. Look at the next figure to see what we are talking about:

      Factorial notation
      Figure 3: Applying the restriction to the letter arrangement

      So, in this case in order to obtain the total amount of different arrangements that can happen when applying the given restriction (having all the vowels together) can be obtained by calculating the different ways in which the six boxes can be arranged, times the different ways in which the vowels can be arranged within their box. Therefore:

      total different arrangements =  6!×3!=(720)(6)=4,320\, 6! \times 3! = (720)(6)=4,320
      Equation 19: Different arrangements of the letters in the word computer when the vowels are kept together

      Where 6! In equation 18 represents the total amount of arrangements of the 6 boxes, and 3! Represents the total amount of arrangements the vowels can have within their own box.

      On the videos for this lesson you will find a few more examples on this topic, so dont forget to watch all of the video lessons.

      ***

      To finalize this lesson we recommend you to visit the next article on factorials which contains a few examples and a different arrangement on how the factorial product is written (remember the order of the factors does not alter the product in a multiplication, and so, a factorial product can be written in different orders).

      This is it for our lesson of today, see you in the next one!