Algebraically Determining the Solution to this Combination

[Euler2718]

Homework Statement

[tex] \dbinom{7}{r} = 21 [/tex]

Homework Equations

[tex] \dbinom{n}{r} = \frac{n!}{(n-r)!r!} [/tex]

The Attempt at a Solution

[tex] \dbinom{7}{r} = 21 [/tex]

[tex] \frac{7!}{(7-r)!r!} = 21 [/tex]

[tex] 7! = 21(7-r)!r! [/tex]

[tex] 240 = (7-r)!r! [/tex]

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)?

 

[ehild]

Homework Statement

[tex] \dbinom{7}{r} = 21 [/tex]

Homework Equations

[tex] \dbinom{n}{r} = \frac{n!}{(n-r)!r!} [/tex]

The Attempt at a Solution

[tex] \dbinom{7}{r} = 21 [/tex]

[tex] \frac{7!}{(7-r)!r!} = 21 [/tex]

[tex] 7! = 21(7-r)!r! [/tex]

[tex] 240 = (7-r)!r! [/tex]

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)?

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.

 

[Euler2718]

2,3,4,5, should cancel with r!

What do you mean?

 

[ehild]

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

Simplifying the fraction, you get 7*3.

 

[Euler2718]

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

Simplifying the fraction, you get 7*3.

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

 

[Mark44]

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

Simplifying the fraction, you get 7*3.

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

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##)?

 

[Euler2718]

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##)?

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?

 

[ehild]

Yes.

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?

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