IB Math topic: 6 Tricky questions involving factorial notation, n!
We come across the Factorial notation while solving the questions of the Binomial Theorem or while learning the solutions of the Permutation and Combination problems. Often we are required to solve one or two questions involving factorials. Therefore, before learning those chapters in detail, let us first of all know what is meant by the factorial notation and then try to solve 6 different types of questions having numbers or variables expressed in factorial notation form.
We define n! = n(n - 1)(n - 2)(n - 3) ... 1
This means if n = 5, then 5! = 5(5 - 1)(5 - 2)(5 - 3)(5 - 4). We stop at (5 - 4) since it is equal to 1 and the last factor of the expansion of the factorial of any number is 1. The expansion is the product of a series of factors, of which the starting factor is the number itself and the successive numbers get reduced by 1 every time till it reaches 1, so the last factor is 1.
Therefore, 5! = 5 * 4 * 3 * 2 * 1, where '*' is the multiplication sign. We multiply the numbers to get the value of the factorial and we say 5! = 120.
Similarly, 4! = 4(4 - 1)(4 - 2)(4 - 3) = 4 * 3 * 2 * 1 = 24
Looking at the expansions of 5! and 4!, we can deduce 5! = 5 * 4!
Again, if we take n = 5, then (n - 1) = 4 and we can generalize the above relation of 5! = 5 * 4! as n! = n * (n - 1)!
5! can also be written as 5 * 4 * 3!, because 3! = 3 * 2 * 1. Therefore, n! = n * (n - 1) * (n - 2)!
Having learnt about these basics of the factorial notation, let us try to solve some questions.
Question 1 Simplify (8!) / (4 * 6!) here '/' means division. 8! is divided by (4 * 6!)
Answer 1 We find 6! in the denominator hence we write 8! as 8 * 7 * 6! so that we can cancel the 6! from both the numerator and the denominator.
(8 * 7 * 6 !) / (4 * 6!) = (8 * 7) / 4 = 2 * 7 = 14. (Always express the larger number's factorial in terms of the smaller number's factorial.)
Question 2 Simplify (4! * 5!) / (3! * 6!)
Answer 2 (4 * 3! * 5!) / (3! * 6 * 5!) = 4/6 = 2/3
Question 3 Simplify (n + 1)! / [n! - (n + 1)!]
Answer 3 Since n is smaller than (n + 1), we express (n + 1)! in terms of n!
Therefore, (n + 1)! / [n! - (n + 1)!] = [(n + 1) * n!] / [n! - (n + 1) * n!]
= [(n + 1) * n!] / [n!{1 - (n + 1)}]
= [(n + 1) * n!] / [n!(-n)]
= - (n + 1) / n
Question 4 Simplify [{(n!)^2} - 1] / (n! - 1)
Answer 4 [{(n!)^2} - 1] / (n! - 1)
= [(n! + 1)(n! - 1)] / (n! - 1) (using the a^2 - b^2 = (a + b)(a - b) identity) (a^2 means 'a' squared or a * a)
= n! + 1
Question 5 Show that [(2n + 2)! {(n!)^2}] / [(2n)! * {(n + 1)!}^2] = [2(2n + 1)]/(n + 1)
Answer 5 We find the numbers (2n + 2) and 2n close to each other and 2n is the smaller number. Therefore, we convert (2n + 2)! to (2n)!. We also convert (n + 1)! to n!.
[(2n + 2)! {(n!)^2}] / [(2n)! * {(n + 1)!}^2]
= [(2n + 2) * (2n + 1) * (2n)! * (n!) * (n!)] / [(2n!) * {(n + 1)!} * {(n + 1)!}]
= [(2n + 2) * (2n + 1) * (2n)! * (n!) * (n!)] / [(2n!) * {(n + 1) * n!} * {(n + 1) * n!}]
= [(2n + 2) * (2n + 1)] / [(n + 1) * (n + 1)]
= [2(n + 1) * (2n + 1)] / [(n + 1) * (n + 1)]
= [2(2n + 1)]/(n + 1)
Question 6 Solve the equation: 16(n - 1)! = 5n! + (n + 1)! where n is any positive integer.
Answer 6 16(n - 1)! = 5n! + (n + 1)!
=> 16(n - 1)! = 5n(n - 1)! + (n + 1)(n)(n - 1)!
=> 16(n - 1)! = (n - 1)![5n + n(n + 1)]
=> 16 = 5n + n^2 + n
=> n^2 + 6n - 16 = 0
=> n^2 + 8n - 2n - 16 = 0
=> n(n + 8) - 2(n + 8) = 0
=> (n + 8)(n - 2) = 0
Therefore, n = - 8 or n = 2. Since n is a positive integer, therefore we have to reject the - 8 value of n. So, the final answer is n = 2.
These are the six sample questions that I have taken up to explain how to solve this type of questions. Please feel free to write to me if you have any question.
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