Do They Divide?

Probability Level 1

Is the following number a whole number?

$\large \frac{90!}{(30!)^3}$

Yes No

4 solutions

Discussions for this problem are now closed

Alexandre Miquilino
Jan 5, 2016

This number is the solution for the following problem:

Consider a string containing $30$ A's, $30$ B's and $30$ C's. How many permutations of this string exist?

We have ninety characters, and three sets of thirty characters that are equal amongst themselves within each set. Thus, we have $90!$ divided by $30!^{3}$ . Since the number of permutations must be an integer, the number $\frac{90!}{(30!)^{3}}$ must be an integer.

Varun Kumar
Jan 5, 2016

$1\cdot 2 \cdot 3 \cdot 4 \cdots 90 = 90!$ has:

3 multiples of 30, ie 30, 60, 90;

3 multiples of 29, ie 29, 58, 87;

3 multiples of 28, ie 28, 56, 84;

3 multiples of 27, ie 27, 54, 81;

3 multiples of 2, ie 2, 4, 6;

3 multiples of 1, ie 1, 2, 3;

Therefore, $\frac{90!}{(30!)^3} = \frac{1\cdot \cdot 3 \cdot 4 \cdots 90}{(1 \cdot 2 \cdot 3 \cdot 4 \cdots 30)^3}$ is a whole number.

There is a flaw in this analysis. You don't show that the factor of 90 counted in one line is distinct from the factor counted in another line, but 90! only has each number once. You count 30 as 1x30, but you also count it as 15x2 and 10x3.

Pete Gast - 4 years, 9 months ago

Trevor B.
Jan 4, 2016

We know that $\dbinom{90}{60}=\dfrac{90!}{60!30!}$ is an integer and a factor of $90!$ . Also, $(30!)^2$ is a factor of $60!$ , so $\dfrac{60!}{30!30!}$ is an integer. Thus, $\dfrac{90!}{60!30!}\times\dfrac{60!}{30!30!}=\dfrac{90!}{30!30!30!}$ is an integer.

Nice solution. You can also use pigeonhole principle to find out the answer. Product of $n$ consecutive positive integers is always divisible by $n!$ .

So $30!$ divides $1 \cdot 2 \cdot 3 \dots 30$ , $30!$ divides $31 \cdot 32 \cdot 33 \dots 60$ and $30!$ divides $61 \cdot 62 \cdot 63 \dots 90$ .

Arulx Z - 5 years, 3 months ago

Can you plz elaborate

Santosh Narva - 5 years, 5 months ago

Another good way to think about this is:

Easy step 1: 90!/(30!)³ = 90x89x88x...x31/(30!x30!)

Step 2:

90x89x88x...x31/(30!x30!) = 90x89x88x...x31/(30x29x28...x2x30!)

We know that the numerator has all integer numbers from 31 to 90, and we know that 58 = 29x2 , 56 = 28x2 , 54 = 27x2

So we can "cancel"all numbers from denominator except 30!

And in the end we have: 90x89x88x...x61x(some numbers here)/(30!)

Step 3:

90x89x88x...61/(30!). Almost like step 2, but a little bit different

90x89x88x...x61/(30!) = 90x89x88x...x61/(30x29x28x27x...x2), and we know that 87 = 29x3 , 84 = 28x3 , 81 = 27x3

So, as we did in Step 2, we can "cancel" every "damn" number who left in denominator, and in the end we would have something like:

90x89x88x86x85x83...(lots of integers here); and we know this is an integer !!

Victor Veras - 5 years, 5 months ago

Nice solution

Utku Demircil - 4 years, 9 months ago

Spend time with Distributing Formula (a basic counting method), you can do it. :)

Krit Phuengphan - 5 years, 5 months ago

Prince Loomba
Jan 20, 2016

Lets divide 90 objects into three groups of 30 each and distribute them into 3 people. The question is the answer to this problem. This would obviously be integer!

Do They Divide?

4 solutions

Discussions for this problem are now closed

0 pending reports