# Algebraically Determining the Solution to this Combination

• Euler2718
In summary, to solve the equation \dbinom{7}{r} = 21 algebraically, you can use the definition of n! and factor out the terms in the numerator and denominator. By simplifying the fraction and considering the factors of 21, you can determine that r=2 or r=5.
Euler2718

## Homework Statement

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

## Homework Equations

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

## 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)?

Morgan Chafe said:

## Homework Statement

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

## Homework Equations

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

## 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)?
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
ehild said:
2,3,4,5, should cancel with r!

What do you mean?

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.

Euler2718
ehild said:
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.

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

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

Morgan Chafe said:
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
Mark44 said:
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?

Yes.
Morgan Chafe said:
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.

Euler2718

## 1. How do you algebraically determine the solution to a combination?

The first step is to identify the variables in the combination and assign them values. Then, use algebraic equations to set up the combination and solve for the desired variable.

## 2. What is the difference between a combination and a permutation?

A combination is a way to select a subset of items from a larger set, without considering the order in which they are selected. A permutation, on the other hand, is a way to select a subset of items with a specific order.

## 3. Can you explain the formula for a combination?

The formula for a combination is nCr = n! / r!(n-r)!, where n represents the total number of items in a set and r represents the number of items being selected from that set. The exclamation mark indicates a factorial, which means multiplying a number by all the positive integers less than it.

## 4. How can I use algebra to solve a combination problem?

To use algebra to solve a combination problem, first set up the combination using the given information and unknown variables. Then, use algebraic equations and properties to manipulate the equation and solve for the unknown variable.

## 5. Are there any tips for solving combination problems quickly?

One helpful tip is to simplify the given information and break down the combination into smaller, more manageable parts. Also, it can be useful to memorize common combination formulas and practice solving problems using them.

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