Question

How many distinct ways to draw a sample of two balls without replacement from a bag containing ten red balls and 4 blue balls.

91
20
40
210

Answer 1

i need to solve that without using calculator

Answer 2

This is a combination problem, not a permutation problem, since order does not matter.

14C2 = (14!)/[(12!)(2!)]

Answer 3

can someone use probability over here ?

Answer 4

Your question is not asking for a probability answer. The problem is a "count". Also, in addition to my above answer:

n! = (n)(n - 1)(n - 2) . . . (3)(2)(1)

Answer 5

The reason the answer is:

14C2

is that distinctness is implying that you look at all 14 balls as unique, not just as red or blue.

Answer 6

So, time for you to answer something. Given what was written above, what is:

14! and 12! and 2!

?

Answer 7

waittt

Answer 8

You don't need a calculator:

14C2 = (14!)/[(12!)(2!)] =

(14 x 13 x 12 x . . . x 3 x 2 x 1)/[(12 x 11 x 10 x . . . x 3 x 2 x 1)(2 x 1)]

(14 x 13)/(2 x 1)

The reason you don't need a calculator is because the factors

(12 x 11 x 10 x . . . x 3 x 2 x 1)

will cancel from the numerator and the denominator.

You are left with:

(14 x 13)/(2 x 1)

Answer 9

All good now, @ArbabShah ?

Answer 10

ohh okay thankyou :D

Answer 11

uw!

Answer 12

Good luck to you in all of your studies and thx for the recognition! @ArbabShah

Answer 13

Thanks man :) hope i just clear my test tomorrow :p

Answer 14

I will hope and think good thoughts for your success!

Answer 15

if we do 10C2+ 10C2 x 4C1 + 4C2

Answer 16

@tcarroll010

Answer 17

When you have anything of the form:

xCy

That implies that x >= y and it resolves down to:

(x!)/[(x-y)!(y!)]

With that formula, you can now do any combination.

Answer 18

can u solve that above example for me 10C2+ 10C2 x 4C1 + 4C2

Answer 19

Can't you make the substitutions?

Answer 20

no confused

Answer 21

Then we probably better work through the confusion. Take that first expression:

10C2

I used formula

xCy

If you place this right underneath, you can tell what "x" and "y" are for sure. Then you can use the other formula I gave you:

(x!)/[(x-y)!(y!)]

And you can definitely do that since you know "x" and "y". And since you also know that:

n! = (n)(n - 1)(n - 2) . . . (3)(2)(1)

you now have, all in one post, all the tools you need to do this problem. You can do this, I have confidence in you.

Answer 22

x=14
y=2?

Answer 23

?

Answer 24

?

Answer 25

x=14
y=2? after this

Answer 26

I guess I just don't understand your question or source of confusion. Sorry.

Answer 27

if you just solve the formula whichever you are using i will be able to understand

Answer 28

That's what I'm trying to get you to do. Your last problem was written:

10C2+ 10C2 x 4C1 + 4C2

In that you have 4 combination expressions. You solve each one of them separately first. Then you can do your addition and multiplication. Each of the 4 combination expressions is identical in format to the problem I already showed you how to do. I gave you formulas and I worked an example in depth. Now it's up to you to use what you have learned and you have everything you need. There is nothing more I can add to this by way of explanation. It's up to you to complete the work now.