Algebraically Determining the Solution to this Combination

1. May 22, 2015

Euler2718

1. The problem statement, all variables and given/known data

$$\dbinom{7}{r} = 21$$

2. Relevant equations

$$\dbinom{n}{r} = \frac{n!}{(n-r)!r!}$$

3. The attempt at a solution

$$\dbinom{7}{r} = 21$$

$$\frac{7!}{(7-r)!r!} = 21$$

$$7! = 21(7-r)!r!$$

$$240 = (7-r)!r!$$

So I get here and the convention is to guess in check (to my knowledge). I found a way to do it on my TI-84; simply go y= 7 nCr x ; then observe the table to gather the solutions of 5 and 2. But how would you do this algebraically (no guess and check)?

2. May 22, 2015

ehild

Write out both the factorials and 21 in product form.
$7! =2\cdot3\cdot4\cdot5\cdot6\cdot7$.
$21=3\cdot7$.
See what factors should cancel.

3. May 22, 2015

Euler2718

What do you mean?

4. May 22, 2015

ehild

Think of the definition of n!.
$$\dbinom{7}{r} = \frac{2\cdot3\cdot4\cdot5\cdot6\cdot7}{(2\cdot3\cdot \cdot\cdot)(\cdot\cdot\cdot)}=3\cdot7$$

Simplifying the fraction, you get 7*3.

5. May 22, 2015

Euler2718

Sorry, I still don't quite follow what's happening in the denominator.

6. May 22, 2015

Staff: Mentor

ehild is asking you to fill in the missing parts, using the definition of $\dbinom{7}{r}$. What should be the last (largest) number in ($2 \cdot 3 \cdot \dots ?$)? What should be the factors in ($\dots$)?

7. May 22, 2015

Euler2718

I think I have some clarity on it. In the numerator, the only pairs of numbers that even have a chance to be divided (once multiplied together) to make 21 are 7 and 6. So that means that 5! must cancel up in the numerator to leave these two factors. By consequence, (7-5)! = 2! = 2, thus 42/2 = 21. It's also then obvious that if r=2, 5! occurs in the (7-r)!, and also works. Does this reasoning make sense?

8. May 23, 2015

ehild

Yes.
Yes. But why do you multiply the factors? 21 = 3*7 and you have to divide 6*7 by 2 to get it.