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January 19th, 2016, 06:30 PM   #1
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In a popular lottery game, five numbers are to be picked randomly from 1 to 36, with no repetitions.

A) how many ways can these five winning numbers be picked without regard to order?

B) answer the same question for picking six numbers.

My attempt:

A) without regard to order, the five winning numbers can be picked in 36C5 = 11088 ways

B) without regard to order, the six winning numbers can be picked in 36C6 = 1947792 ways

Am I correct? Is there another way of solving this? Thanks.
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January 19th, 2016, 07:18 PM   #2
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Looks good.
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January 20th, 2016, 03:41 AM   #3
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I am not sure what you mean by "another way of solving this" since you have not told how you solved it. Did you just apply a formula that says that "the number of ways n things can be picked out of m without regard to order is nCm"?

I would argue that, out of 36 numbers, there are 36 ways to choose one. Then, since there are no repetitions, there are 35 ways to choose the next, 34 for the third, 33 for the fourth, and 32 for the fifth so 36*35*34*33*32 different numbers and orders. But there are 5!= 120 ways to order 5 things so the number of different selections, without regard for order, are (36*35*34*33*32)/120= 376992 ways to select 5 things out of 36 without regard for order.
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January 20th, 2016, 04:15 AM   #4
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Country Boy's select is exactly ${36 \choose 5}$, but he has evaluated the answer correctly.

Shunya has the second one correct.
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January 20th, 2016, 06:58 AM   #5
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Country Boy and v8archie

Thanks for checking my understanding. I'm grateful to you both.

Please see below if I've understood things or not. I will use concrete examples as they are easier to handle.

How many different 3-digit numbers can be formed with the digits 4,6,9 given that the digits don't repeat?

With numbers order matters 469 is not the same as 946.

We can consider _ _ _ to be the three positions of a 3-digit number.
So...
There are 3 ways to fill the first position. For each of these here are 2 remaining digits for the second position, giving us 3*2 = 6 possible ordered arrangements. For each of these there's 1 (the last remaining digit) way to fill the third position. This gives us 3*2*1 = 6 different 3-digit numbers disallowing repetition.

I have some difficulty wrapping my head around combination. Anyway let me show you how much I understand it with an example.

From a group of 10 people how many 5-member basketball teams can be formed?

In this case (forming a team) order doesn't matter. So...

Let x be the number of 5-member teams that can be formed out of a group of 10 people.

Each of these ''combinations'' can be arranged in ''order'' in 5*4*3*2*1 = 5! different ways.

So for all ''combinations'' there's x*5! different ''ordered'' arrangements (permutation''

But x*5! = 10*9*8*7*6
So, x = (10*9*8*7*6)/5! = 10P5/5! = 10C5

That's the way I understand it. Am I making any sense? Thanks.
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