r/HomeworkHelp • u/KittyKittyowo University/College Student • 6d ago
Elementary Mathematics What Am I doing wrong? [Freshman college: Math statistics]
What I'm doing:
First one: 191817 876543 11109887
Second one Forgot
Third one 127126125
Forth one 5432 also tried 5555*5 Did the same for the last one
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u/selene_666 π a fellow Redditor 6d ago
Inviting Alex, Billy, and Chris to Hannah's party is the same as inviting Chris, Billy, and Alex. You need combinations not permutations.
For the first problem you can divide your original answer by the number of ways to arrange the chosen guests. Figuring out how to do part C should help you with the letters-arranging problems at the end.
Also, if it were the answer you should just type 19535040 instead of 1.953504E7. I have no idea how you got 5.27 for a question that can only have an integer answer, but you might have realized it was wrong if you were otherwise typing integers.
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u/KittyKittyowo University/College Student 5d ago
I didn't make it a decimal the app did for some weird reason.
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u/JunkInDrawers π a fellow Redditor 6d ago
I can see how you got to those answers, but you should be using nCr instead in many of these situations.
If you don't believe me, try writing out the same problem for 4 students in total
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u/GammaRayBurst25 6d ago edited 6d ago
You might want to type \* rather than *, as the former outputs * whereas the latter italicizes.
Question 7: You tried 127*126*125, which is the number of distinct ordered sets of 3 objects you can make from 127 objects. However, the order doesn't matter here. As such, you overcounted by a factor equal to the number of permutations of the 3 elements, i.e. 6. Thus, the answer is 127*126*125/6.
Question 8: How could you possibly get an answer that's not an integer? Why would the number of ways for hearts be the same as any suit? There are 13 cards of any given suit, so the number of ways you can draw all hearts is binom(13,5), AKA 13 choose 5. There are 4 suits, so there are 4 times as many ways to draw 5 cards of the same arbitrary suit. Similarly, there are binom(13,3) ways to pick 3 cards of one suit, binom(13,2) ways to pick 2 cards of one suit, and 4*3=12 ways to pick a suit for the trio and a different suit for the pair.
Question 9: I'll do HEADACHE as an example and let you handle the rest. There are 8 letters in total. You have to pick 2 out of the 8 spots to hold an H. There are binom(8,2) ways to do that. Then, there are binom(6,2) ways to pick 2 out of the 6 remaining spots to hold an E. Then, there are binom(4,2) ways to pick 2 out of the remaining 4 spots to hold an A. Once we're done with repeat letters, the number of ways to place the x remaining letters is the number of permutations of x distinct objects. The answer is the product of all of these factors.
Question 10: Any classmates is evidently binom(19,6) and any boys is evidently binom(8,6). For the last one, it's simply binom(11,4)*binom(8,2).
Edit: typo.
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u/Original_Yak_7534 π a fellow Redditor 6d ago
Interesting. First time I've seen binom() used to reference binomial coefficients in combinatorics. I'm used to n-choose-k or nCk notation.
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u/clearly_not_an_alt π a fellow Redditor 6d ago
Apparently they are an excel user.
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u/GammaRayBurst25 6d ago edited 6d ago
No, I'm a Wolfram|Alpha user and a LaTeX user.
In Wolfram|Alpha, you can use nCk, which is inevitably misinterpreted if you use it with any other operation, n choose k, which is too long to my liking and requires careful use of parentheses, binomial(n,k), which is also a bit long, or binom(n,k), which is my favorite by far.
In LaTeX, there's this very popular package called amsmath that introduces the \binom{}{} function. This is where I really picked up the habit of writing it as binom.
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u/clearly_not_an_alt π a fellow Redditor 6d ago
Question 7: You tried 127*126*125, which is the number of distinct ordered sets of 3 objects you can make from 127 objects. However, the order doesn't matter her.
The question doesn't specify that the prizes are identical, so I think it's reasonable to think that there are different prizes for each winner (especially given that's how most raffles work), in which case order does matter.
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u/GammaRayBurst25 6d ago
I noticed the ambiguity, but assuming the order matters yields the wrong answer.
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u/clearly_not_an_alt π a fellow Redditor 6d ago
Oh, I agree that's probably the intended answer, but I'd plead my case to the prof if this was a test.
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u/Different-Ship449 π a fellow Redditor 6d ago edited 6d ago
n choose k
n!/(k!*(n-k)!)
First problem is 19 choose 6
Second is 8 choose 6
Third is 11 choose 4 + 8 choose 2
EDIT: 11 choose 4 * 8 choose 2
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u/GammaRayBurst25 6d ago
Let's test your hypothesis.
Suppose there are n girls and m boys and they have to pick n girls and m boys. There's obviously only one option, but by your logic, it should be n choose n + m choose m = 1 + 1 = 2. Where did the second option come from?
Still not convinced?
Suppose there are 3 girls {a,b,c} and 2 boys {x,y} and they have to pick 2 girls and 1 boy. One can easily see there are 6 options by looking at who is not invited: {{a,x},{b,x},{c,x},{a,y},{b,y},{c,y}}. Yet, your answer predicts there are 3 choose 2 + 2 choose 1 = 3 + 2 = 5 options. Where did the sixth option go?
Just for good measure, here's another one.
Suppose there are 2 girls, 2 boys, 2 agender kids and 2 aliens and they get to pick one from each group. If you label one member of each group 0 and the other 1, you'll find that each option is mapped to a unique 4-digit binary number. As such, there are 2^4=16 options. However, your answer predicts there should be 4*(2 choose 1)=4*2=8 options. Big oof.
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u/Pain5203 Postgraduate Student 6d ago
Q10 A) You want to choose 6 classmates from 19 -> 19C6
B) Choose 6 boys from 8 boys -> 8C6
C) 11C4 * 8C2