Combinatorics: Factorials- Basic Definitions

If $$x$$ is a positive integer greater than 2, the product of $$x$$ consecutive positive integers must be divisible by which of the following? >I. $$x- 1$$ >II. $$2x$$ >III. $$x!$$

At this point, you can choose answer choice E with a high degree of certainty that all of the statements must be true. If you have the time, take a look at the algebra and prove why the statements must be true: The product of $$x$$ consecutive integers is divisible by $$x!$$, so statement III is definitely true. Now, $$x!$$ is >$$1\times 2\times 3\times$$ … $$(x-1)\times x $$, so $$(x-1)$$ and 2 are included in $$x!$$ (that is, provided that $$x$$ is greater than 2). Hence, this is the correct answer.

Incorrect. [[snippet]] Yes, statements II and III are true. But what about statement I? Can you find an example eliminating this statement?

Incorrect. [[snippet]] Yes, statement I and III are true. But what about statement II? Can you find an example eliminating this statement?

Incorrect. [[snippet]] Yes, statement I is true. But what about statements II and III? Can you find an example eliminating these statements?

Incorrect. [[snippet]] Yes, statement II is true. But what about statements I and III? Can you find an example eliminating these statements?

Correct. [[snippet]] __Plug In__ $$x = 3$$. In this case, the product could be $$3 \times 4 \times 5 = 60$$. >I. 60 is divisible by $$x - 1 = 3 - 1 = 2$$. >II. 60 is divisible by $$2x = 2 (3) = 6$$. >III. 60 is divisible by $$x! = 3! = 3 \times 2 \times 1 = 6$$. __Plug In__ another value of $$x$$ to be sure because the question says *must be*. This time try an even integer (such as $$x= 4$$). In this case, the product could be $$2 \times 3 \times 4 \times 5 = 120$$. >I. 120 is divisible by $$x - 1 = 4 - 1 = 3$$. >II. 120 is divisible by $$2x = 2 (4) = 8$$. >III. 120 is divisible by $$x! = 4! = 4 \times 3 \times 2 \times 1 = 24$$.

I only

II only

II and III only

I and III only

I, II, and III

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