Factorials & divisibility

[binaras]

Hi

need some help with the following

Question
Which of the following is an integer
1. 12!/6!
2. 12!/8!
3. 12!/7!5!

esp with the 3.12!/7!5!

Thanks

[DavidG@VeritasPrep]

Let's write these expressions out. If we can completely cancel out the denominator, then we'll be left with an integer value.

1) (12*11*10*9*8*7*6*5*4*3*2*1)/(6*5*4*3*2*1)
We can cancel out a 6! in both numerator and denominator, which leaves us with 12*11*10*9*8*7. No need to calculate. Clearly an integer.

2) (12*11*10*9*8*7*6*5*4*3*2*1)/(8*7*6*5*4*3*2*1)
We can cancel out an 8! in both numerator and denominator, which leaves us with 12*11*10*9. Again, no need to calculate. Clearly an integer.

3)(12*11*10*9*8*7*6*5*4*3*2*1)/[(7*6*5*4*3*2*1)*(5*4*3*2*1)]
We can cancel out a 7! in both numerator and denominator, which leaves us with (12*11*10*9*8)/(5*4*3*2*1)

The 5 and the 2 in the denominator will cancel out the 10 in the numerator

Now we have (12*11*9*8)/(4*3*1)

The 4 and the 3 in the denominator will cancel out the 12 in the numerator

We're left with 11*9*8. Clearly an integer.

[[email protected]]

Hi binaras,

Beyond the arithmetic that David explained, these calculations fall into some patterns that you are likely to see (in some form) on Test Day.

12!/6!

When dealing with INDIVIDUAL factorials, if the factorial in the numerator is GREATER than the factorial in the denominator, then you will end up with an integer. Here, since 12 > 6, 12!/6! will simplify to an integer.

12!/7!5!

You might not have studied the Combination Formula yet, but this calculation is what you would end up with when answering the question "how many different combinations of 7 people can you form from a group of 12 people?" In these types of "Combination" calculations, the number of groups will always be an integer, so 12!/7!5! will simplify to an integer.

GMAT assassins aren't born, they're made,
Rich

[binaras]

thanks

[Jeff@TargetTestPrep]

Hi

need some help with the following

Question
Which of the following is an integer
1. 12!/6!
2. 12!/8!
3. 12!/7!5!

esp with the 3.12!/7!5!

Thanks

Solution:

Before actually solving this problem, let's review how factorials can be expanded and expressed. As as example, we can use 5!.

5! could be expressed as:

5!

5 x 4!

5 x 4 x 3!

5 x 4 x 3 x 2!

5 x 4 x 3 x 2 x 1!

Understanding how this factorial expansion works will help us work our way through each answer choice, especially answer choices 1 and 2.

1. 12!/6!

Since we know that factorials can be expanded, we now know that:

12! = 12 x 11 x 10 x 9 x 8 x 7 x 6!

Plugging this in for answer choice 1, we have:

(12 x 11 x 10 x 9 x 8 x 7 x 6!)/6! = 12 x 11 x 10 x 9 x 8 x 7, which is an integer.

2. 12!/8!

Once again, since we know that factorials can be expanded, we now know that:

12! = 12 x 11 x 10 x 9 x 8!

Plugging this in for answer choice 2, we have:

(12 x 11 x 10 x 9 x 8!)/8! = 12 x 11 x 10 x 9, which is an integer.

3. 12!/(7!5!)

Once again, since we know that factorials can be expanded, we now know that:

12! = 12 x 11 x 10 x 9 x 8 x 7!

Plugging this in for answer choice 3 gives us:

(12 x 11 x 10 x 9 x 8 x 7!)/(7!5!)

(12 x 11 x 10 x 9 x 8)/(5 x 4 x 3 x 2 x 1)

(12 x 11 x 10 x 9 x 8)/(12 x 10 x 1)

11 x 9 x 8, which is an integer.

We see that choices 1, 2, and 3 are all integers.