Lesson: Calculating Combinations

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Why and how is 0!=1?
gmat-admin's picture

Keep in mind that n! = the product of POSITIVE integers from n to 1. So, 0! doesn't even make any sense, since factorials are only defined for positive integers.

That said, we need to have some agreed-upon convention for dealing with 0! so that we can ensure that calculations involving factorials make sense.

For example, the number of ways to select 3 people from 3 people = 3C3 = 3!/(3!)(0!)

We already know that there is only 1 way to select 3 people from 3 people. So, we need 0! to equal 1. As such, this is an agreed-upon convention.

In this case since "though not necessarily right behind him" shouldn't we,
360 - (# of time J is behind F = 120 ) == 240 ?
Please help.
gmat-admin's picture

Question link: http://www.beatthegmat.com/mobster-combinatorics-t166632.html

The words "not necessarily right behind him" do not change the answer.

The key concept here is that, if we IGNORE the restriction, we can arrange the 6 people in 720 ways.

Of these 720 different arrangements, Frankie will be standing behind Joey half the time, and Joey will be standing behind Frankie half the time. (in other words, for every arrangement where Frankie is ahead of Joey, there is an arrangement where Joey is ahead of Frankie).

Does that help?


Hi Brent,

Could you help me with the below question please.

How many 3 digit integers can be chosen such that none of the digits appear more than twice and none of the digits equal zero ?

A. 729
B. 720
C. 648
D. 640
E. 576

I solved it as 9 * 9* 8 and got it wrong.

gmat-admin's picture

I think you're missing the phrase "more than TWICE." This means it's okay to have a 3-digit number that has a digit appear TWICE, but it's not okay to have a digit appear THREE TIMES. So, for example, the numbers 664 and 922 are okay, but the number 777 is not okay.

The quickest solution is to first determine the number of ways to create ANY 3-digit number that does not contain any zeros.

For each digit, we have 9 options (1,2,3,4,5,6,7,8 or 9).
So, the TOTAL number of outcomes = (9)(9)(9) = 729

Of course, some of these 729 outcomes break rule about not allowing a digit to appear THREE TIMES. So, we must subtract from 729 all of the outcomes that break the rule.

Fortunately, there aren't many rule-breakers among the 729 outcomes.

The rule-breakers are the following numbers: 111, 222, 333, 444, 555, 666, 777, 888 and 999 (9 altogether).

So, the number of outcomes that satisfy the given restrictions = 729 - 9
= 720

Does that help?


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