<p>The buyer of a lottery ticket chooses 4 numbers from 1 to 32. Repetition is not allowed.
How many combinations of 4 numbers are possible?</p>
<p>How do I do this…?</p>
<p>is it 32 x 31 x 30 x 29 or 32C4?</p>
<p>32C4 </p>
<p>10char</p>
<p>It depends on whether or not the order of the numbers matters. Now it seems like the order of the numbers on a lottery would in fact matter, but the problem says “how many combinations”, so I’m not sure. (if the order matters its 32 P 4 and if it doesn’t then 32 C 4)</p>
<p>I can’t think of a lottery drawing where order matters. You guess your numbers and then they draw the ping-pong balls. You don’t have to guess which order the balls will be drawn — they list the winning numbers in ascending order but they don’t have to be drawn that way!</p>
<p>There’s 32 choices for the first one, 31 for the second, 30 for the third, and 29 for the fourth. So 32 x 31 x 30 x 29.</p>
<p>If order matters, then it’s (32 x 31 x 30 x 29) x 4!. If not, it’s just 32 x 31 x 30 x 29.</p>
<p>There’s some good lessons on counting techniques at [SAT</a> Prep help - ViaTutor.com](<a href=“http://www.viatutor.com%5DSAT”>http://www.viatutor.com). Here’s a link to a lesson, if you’re having trouble: [SAT</a> Math resources](<a href=“viatutor.com”>viatutor.com)</p>
<p>^^^^</p>
<p>It’s 32x31x30x29 if order DOES matter (that is: 1,2,3,4 is a different answer than 4,3,2,1)</p>
<p>and then 32x31x30x29/4! if order DOES NOT matter…</p>
<p>I guess I’m just too unfamiliar with the lottery system 
Well then the answer would just be 32 C 4 - and pckeller is correct in that akorch was incorrect.</p>
<p>pckeller has the right answer – sorry about that!</p>
<p>its definitely 32 31 30 29 because its essentially just asking you how many different arrangements can you make, and this is how you would get it. since you can’t repeat you have 32 options than 31 than 30 than 29. not sure if everyone was already agreed on that. its the fundamental counting principle isn’t it?</p>
<p>But it is NOT an arrangement question – it’s a combination question. Just think: have you ever heard some one crying – “I picked ALL of the right lottery numbers, but I got them in the wrong order!”</p>
<p>Say you picked 2, 4, 6 and 8. And they drew the winning numbers: 2, 4, 6 and 8. Obviously, you win. But if instead, they drew 8,4,2,6…you still win. Order does not matter.</p>
<p>so i am right? i dont understand. What the question is asking is do you know how to identify how many options are available if you have this 32 with 4 slots and repeats are not allowed. order doesn’t matter but if you pick a number first it cant be repeated again so the option of choosing that number is eliminated and there are only 31 choices left.</p>
<p>you wouldnt divide by four. that doesn’t make any sense.</p>
<p>You don’t divide by 4 — you divide by 4! which is 24.</p>
<p>Here’s why:</p>
<p>Again, suppose you picked the numbers 2, 4, 6, 8. If the lottery reqired you to have them in the right order, then the only way you win is if the numbers drawn are 2, 4, 6 and 8 IN THAT EXACT ORDER. But since lotteries DON’T work that way, you win in any of these cases:</p>
<p>2 4 6 8
2 4 8 6
2 6 4 8
2 6 8 4
2 8 4 6
2 8 6 4
4 2 8 6
4 2 6 8
and so on…</p>
<p>there are 4 x 3 x 2 x 1 ways to arrange these numbers = 4! = 24</p>
<p>So your original calculation: 32 x 31x 30 x 29 has overstated the number of possibilities by a factor of 24. That’s why we divide.</p>
<p>And 32x31x30x29/4! is exactly what your calculater does when you enter 32C4.</p>
<p>Hope that helps…</p>
<p>but its not asking how many different ways you can win. its asking you how many different arrangements of numbers you can make i think. so even if you get 2468. 4268 is another arrangement.</p>
<p>Nope, read it again: it asks how many COMBINATIONS.</p>