# How to answer a Combinations question

• I
Question: When playing Lotto 649, you must pick six numbers from the numbers 1, 2, …, 49. In how many ways can you do this?

My attempt

n!/r!(n-r)! = 49!/6!(49-6)!
(49x48x47x46x.....x1)/(6x5x4x3x2x1)(43x42x41x40x.....x1)

=0.06 .
(edit: i redid the question and just made it (49x48x47x46x45x44)/(6!) and i got 13,983,816. this seems like it could be the right answer but i am still hesitant because im not sure that is legal to just remove the 43! from the denominator)

this is definitely wrong, since the number should be huge, but im not sure how to approach the correct answer

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jedishrfu
Mentor
I think your formula is missing some parens:

## n! / r! (n-r)! ## which is in fact ##\frac {n!} {r!} (n-r)! ## should be ##\frac {n! } {(r! (n-r)!)} ##

by way of explanation, you choose 6 numbers out of 49 which is 49*48*47*46*45*44 or
49! / (49-6)! and now since the order doesn't matter you divide by 6!

Alright, so if i did the equation

(49*48*47*46*45*44)/(6*5*4*3*2*1) ,
so therefore the answer would be 13,983,816 ?

jedishrfu
Mentor
Just understand the meaning of your formula ie basically learn how to derive it and understand how it works and what its limitations are...

Walk around and amaze your friends by computing probabilities of related events.

The mistake you made here is a common meme among folks studying the order of math operations in fuzzy expressions that's why I added the parens to make it clear.

https://www.insider.com/hard-viral-...#this-viral-math-question-has-two-solutions-1
In particular, look at the 6/2(1+2) example. People really get confused by it with some thinking its
##\frac {6} {2} (1+2) = 9## versus some thinking its ## \frac {6} { 2 * (1+2)} = 1##

Do you see the confusion and why the first one is how we interpret the expression with modern PEMDAS conventions?