Which of the following correctly simplifies 10!/8! without expanding either factorial in full?
A10!/8! = 10 × 9 × 8 = 720
B10!/8! = 10 × 9 = 90
C10!/8! = 2! = 2
D10!/8! cannot be simplified without a calculator because both factorials must be computed first
When dividing factorials, factors common to both numerator and denominator cancel. 10! = 10 × 9 × 8 × 7 × ... × 1 and 8! = 8 × 7 × ... × 1. Everything from 8 down to 1 appears in both and cancels, leaving only 10 × 9 = 90. This cancellation pattern is fundamental to combinatorics: permutation and combination formulas always reduce to a short product of consecutive integers, never requiring you to compute the full factorial values.
Question 2 Multiple Choice
A student claims that 0! = 0 because 'the product of no numbers is zero.' What is wrong with this reasoning?
AThe student is right — 0! = 0 follows from the same logic that 0 × anything equals 0
B0! is undefined because you cannot meaningfully multiply zero numbers together
C0! = 1 because the empty product equals 1 by mathematical convention, and this is required for combinatorial formulas like C(n,0) = 1 to remain consistent
D0! = 1 is purely a convenient definition with no deeper mathematical justification
The 'product of no numbers' is not zero — it is 1. This is the empty product convention: multiplying zero factors together gives the multiplicative identity (1), just as summing zero terms gives the additive identity (0). There are also two independent justifications: first, C(n,0) = n!/(0! · n!) must equal 1 (there is exactly one way to choose nothing), which requires 0! = 1; second, working backward from n! = n × (n-1)! gives 1!/1 = 0!, so 0! = 1. If 0! = 0, every combinatorial formula involving 'choosing zero' would be undefined or wrong.
Question 3 True / False
For large values of n, n! grows much faster than 2^n.
TTrue
FFalse
Answer: True
Factorial growth is superexponential. At n=10: 2^10 = 1,024 while 10! = 3,628,800. At n=20: 2^20 ≈ 1,000,000 while 20! ≈ 2.4 × 10^18. The reason is that 2^n multiplies by a constant factor (2) at each step, while n! multiplies by an increasing factor (n) at each step. Once n > 2, each factorial step multiplies by more than the exponential step, and the gap widens rapidly. This explosive growth is why 52! (shufflings of a card deck) is astronomically large — roughly 8 × 10^67.
Question 4 True / False
0! = 0 because zero factorial means multiplying zero copies of a number, which produces zero.
TTrue
FFalse
Answer: False
This is the most common misconception about factorials. The 'product of no numbers' is 1 (the empty product), not 0. Intuition: there is exactly one way to arrange zero objects — the empty arrangement — so the count of arrangements is 1, matching 0! = 1. Algebraically, working backward from 1! = 1 using the rule n! = n × (n-1)! gives 1!/1 = 0! = 1. Practically: setting 0! = 0 would break C(n,0) = n!/(0! × n!) = 1, a formula that must equal 1 for every n.
Question 5 Short Answer
Explain why the number of ways to arrange n distinct objects in a sequence equals n!, using the multiplication principle.
Think about your answer, then reveal below.
Model answer: When placing n distinct objects in a row, you have n choices for the first position. Once placed, n−1 objects remain, giving n−1 choices for the second position. Continuing this way: n−2 choices for the third, n−3 for the fourth, and so on down to exactly 1 choice for the final position. The multiplication principle states that the total number of distinct sequences equals the product of the choices at each independent step: n × (n−1) × (n−2) × ... × 2 × 1 = n!. Each multiplication reflects the fact that every choice at one position combines with every possible choice at all other positions.
This derivation reveals why n! is more than just a notation shorthand — it is the direct answer to a concrete counting problem about arrangements. Every application of factorials in permutations and combinations traces back to this same reasoning about how many independent choices exist at each sequential step.