- #1

mr_coffee

- 1,629

- 1

Wow, I'm totally lost on where to start for this one.

In this chapter we have been working with r-combinations with repeition allowed and using the form of(r + n - 1)

(r )

Where n stands for categories, so if you had 4 categories u would use 3 bars to break up the categories. And r chocies.

For example, I think I did a simplier problem that might help me:

Find how many solutions there are to given equation that satisfy the given condition.y_1 + y_2 + y_3 + y_4 = 30, each y_i is an integer that is at least 2.

So y_i >= 2.

There are 4, y's, so you have 4 categories, thus 3 bars to split up the categories. I will represent the sum by putting x's in each category so the end result equals 30, this will help me visualize the combinations.

There are already 2 x's in each category to start with before I even begin to place any x's in, so that means 30-8 = 22. So i have 22 x's i can distribute.

xxxxxxxxxx | xxxxxxxxxx | x | x

I have 22 x's here, + 8 already in there = 30.

So apply the formula, I have (22 + (4-1)) choose (22)

= 25 choose 22

= 25!/(22!3!) = 2300.Now can I apply this technique to the above problem? If so I really don't see how...

I could do like

x_1 + x_2 + x_3 +...+ x_99999 = 9

but that means I would have 99,999 - 1 categories, and 9 x's to distrubute in those 99,998 categories anywhere I want i guess.

So with that logic the combination would look like:

(9 + 9998) choose 9

= (10007) choose 9

= (10007)!/[9!(9998)!]

= undefined by the all mighty ti-89

which can't be right...any idea where I'm messing up?

Thanks

In this chapter we have been working with r-combinations with repeition allowed and using the form of(r + n - 1)

(r )

Where n stands for categories, so if you had 4 categories u would use 3 bars to break up the categories. And r chocies.

For example, I think I did a simplier problem that might help me:

Find how many solutions there are to given equation that satisfy the given condition.y_1 + y_2 + y_3 + y_4 = 30, each y_i is an integer that is at least 2.

So y_i >= 2.

There are 4, y's, so you have 4 categories, thus 3 bars to split up the categories. I will represent the sum by putting x's in each category so the end result equals 30, this will help me visualize the combinations.

There are already 2 x's in each category to start with before I even begin to place any x's in, so that means 30-8 = 22. So i have 22 x's i can distribute.

xxxxxxxxxx | xxxxxxxxxx | x | x

I have 22 x's here, + 8 already in there = 30.

So apply the formula, I have (22 + (4-1)) choose (22)

= 25 choose 22

= 25!/(22!3!) = 2300.Now can I apply this technique to the above problem? If so I really don't see how...

I could do like

x_1 + x_2 + x_3 +...+ x_99999 = 9

but that means I would have 99,999 - 1 categories, and 9 x's to distrubute in those 99,998 categories anywhere I want i guess.

So with that logic the combination would look like:

(9 + 9998) choose 9

= (10007) choose 9

= (10007)!/[9!(9998)!]

= undefined by the all mighty ti-89

which can't be right...any idea where I'm messing up?

Thanks

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