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mckriddle

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Sep 8, 2013
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Having trouble understanding how to plug in this formula...

There are 52 colored balls in a large tumbler, 13
red, 13 blue, 13 yellow, and 13 green. The balls
of each color are lettered A through M. Five
balls are chosen at random.

How many 5-ball choices consist of exactly 4 balls with the same letter?

 
Hello, mckriddle!

Having trouble understanding how to plug in this formula . . . What formula?

There are 52 colored balls in a box: 13 red, 13 blue, 13 yellow, and 13 green.
The balls of each color are lettered A through M.
Five balls are chosen at random.

How many five-ball choices consist of exactly 4 balls with the same letter?
Click to expand...

We want a set of 4 balls with the same letter.

There are \(\displaystyle \color{blue}{13}\) choices for the letter.

There is \(\displaystyle \color{blue}{1}\) way to get the four balls with that letter.

There are \(\displaystyle \color{blue}{48}\) choices for the fifth ball.


Therefore: .\(\displaystyle 13\cdot1\cdot48 \:=\:\color{blue}{624}\) five-ball choices.
 
That's the formula I was speaking of, I suppose.. the 13*1*48.

I've crammed so many new ways to find answers that I can't quite see it clearly anymore, I guess.
Thanks for your help!
 
mckriddle said:
That's the formula I was speaking of, I suppose.. the 13*1*48.

I've crammed so many new ways to find answers that I can't quite see it clearly anymore, I guess.
Thanks for your help!
Click to expand...
Quite often you don't NEED a formula - just use common sense. Look more at the reasoning Soroban used.
 
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