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Sammy has flavors of candies with which to make goody bags
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Updated on: 16 Aug 2012, 23:12
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Sammy has x flavors of candies with which to make goody bags for Franks birthday party. Sammy tosses out y flavors, because he doesnt like them. How many different 10flavor bags can Sammy make from the remaining flavors? (It doesnt matter how many candies are in a bag, only how many flavors). (1) If Sammy had thrown away 2 additional flavors of candy, he could have made exactly 3,003 different 10flavor bags. (2) x = y + 17 Sourcejamboree
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Originally posted by Marcab on 16 Aug 2012, 22:53.
Last edited by Bunuel on 16 Aug 2012, 23:12, edited 1 time in total.
Edited the question.




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Re: Sammy has flavors of candies with which to make goody bags
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16 Aug 2012, 23:32
Sammy has x flavors of candies with which to make goody bags for Franks birthday party. Sammy tosses out y flavors, because he doesnt like them. How many different 10flavor bags can Sammy make from the remaining flavors? (It doesnt matter how many candies are in a bag, only how many flavors).In order to calculate how many 10flavor bags can Sammy make from the remaining (xy) flavors, we should know the value of xy. The answer would simply be \(C^{10}_{xy}\). For example if he has 11 flavors (if xy=11), then he can make \(C^{10}_{11}=11\) different 10flavor bags. (1) If Sammy had thrown away 2 additional flavors of candy, he could have made exactly 3,003 different 10flavor bags. We are told that \(C^{10}_n=3,003\), where \(n=(xy)2\): he can make 3,003 10flavor bags out of n flavors. Now, n can take only one particular value, so we can find n (it really doesn't matter what is the value n, important is that we can find it), hence we can find the value of xy (xy=n+2). Sufficient. (2) x = y + 17 > xy=17. Directly gives us the value of xy. Sufficient. Answer: D. Hope it's clear.
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Re: Sammy has flavors of candies with which to make goody bags
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17 Aug 2012, 01:17
Hey thanx. But I shall like to bother you again. I have come across around 3 questions of this type where the value of the total, of which we are supposed to make selection, is not known. Even in this question also, I knew that there could be only one value for which the answer comes out to be 3003, but since I was unable to come out with the solution, I sought help. So my question is that can't there be any more value for n for which nC10 is 3003?
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Re: Sammy has flavors of candies with which to make goody bags
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17 Aug 2012, 01:30



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Re: Sammy has flavors of candies with which to make goody bags
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30 Jun 2013, 23:46



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Re: Sammy has flavors of candies with which to make goody bags
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29 Dec 2013, 15:35
Marcab wrote: Sammy has x flavors of candies with which to make goody bags for Franks birthday party. Sammy tosses out y flavors, because he doesnt like them. How many different 10flavor bags can Sammy make from the remaining flavors? (It doesnt matter how many candies are in a bag, only how many flavors).
(1) If Sammy had thrown away 2 additional flavors of candy, he could have made exactly 3,003 different 10flavor bags. (2) x = y + 17
Sourcejamboree OK, let me try this one Question stem basically says we have xy flavors and we need to pick 10 out of these How many combinations can we make? Statement 1 So this is giving us the number of combinations of xy2, therefore we can imply what xy is. Do we need the calculation? No, there will only be one number that will give this answer when deciding to pick 10 out of it. Suff Statement 2 Rearranging xy = 17 We have the total Suff Answer is D Hope it helps Cheers! J



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Re: Sammy has flavors of candies with which to make goody bags
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08 Mar 2016, 18:19
I was concerned more with finding the value of y..but looks like it was not really necessary... how i approached the question: 1. (xy2)C10 = 3003 > find prime factorization of 3003...we can see that we have 3*7*11*13... (xy2)x(xy1)x(xy)(xy+1) x etc. / 10!*(xy12)!
how can we solve further???
2. xy=17 17C10  we can find the answer... so somewhere I did not see how to solve A...



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Re: Sammy has flavors of candies with which to make goody bags
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15 Jul 2016, 09:44
Can anyone help how to solve the combination: (xy2)C10=3003



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Re: Sammy has flavors of candies with which to make goody bags
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30 Jul 2017, 16:48
Marcab wrote: Sammy has x flavors of candies with which to make goody bags for Franks birthday party. Sammy tosses out y flavors, because he doesnt like them. How many different 10flavor bags can Sammy make from the remaining flavors? (It doesnt matter how many candies are in a bag, only how many flavors).
(1) If Sammy had thrown away 2 additional flavors of candy, he could have made exactly 3,003 different 10flavor bags. (2) x = y + 17 We need to determine the number of different 10flavor goody bags Sammy can make for Frank. Since there were x flavors to begin with and y were thrown away, there are (x  y) remaining flavors. Moreover, since the order in which the flavors are chosen is not important, this is a combination problem. In other words, we must determine whether we have enough information to calculate (x  y)C10. Statement One Alone: If Sammy had thrown away 2 additional flavors of candy, he could have made exactly 3,003 different 10flavor bags. This statement tells us that (x  y b 2)C10 = 3003. The important thing to remember when approaching questions like this is that we don’t actually need to find the answer; we simply need to determine whether we have enough information to answer the question. In this case, since there is a unique value for (x  y  2) that satisfies (x  y  2)C10 = (x  y  2)!/((x  y  12)! * 10!), we can determine (x  y  2), and thus we can determine x  y. After x  y is determined, it is easy to calculate (x  y)C10. Statement one alone provides enough information to answer the question. Statement Two Alone: x = y + 17 We have x  y = 17; thus, 17C10 = 17!/(7! * 10!) can be calculated. Statement two alone provides enough information to answer the question. Answer: D
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Sammy has flavors of candies with which to make goody bags
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10 Jul 2018, 04:32
How do you make sure that it has only one solution without solving it, did we just assume it? It may have 10 different roots. (let's say xy2 is a) C(a, 10) = ((a).(a1).(a2)....(a9))/10!=3003 Bunuel wrote: Sammy has x flavors of candies with which to make goody bags for Franks birthday party. Sammy tosses out y flavors, because he doesnt like them. How many different 10flavor bags can Sammy make from the remaining flavors? (It doesnt matter how many candies are in a bag, only how many flavors).
In order to calculate how many 10flavor bags can Sammy make from the remaining (xy) flavors, we should know the value of xy. The answer would simply be \(C^{10}_{xy}\). For example if he has 11 flavors (if xy=11), then he can make \(C^{10}_{11}=11\) different 10flavor bags.
(1) If Sammy had thrown away 2 additional flavors of candy, he could have made exactly 3,003 different 10flavor bags. We are told that \(C^{10}_n=3,003\), where \(n=(xy)2\): he can make 3,003 10flavor bags out of n flavors. Now, n can take only one particular value, so we can find n (it really doesn't matter what is the value n, important is that we can find it), hence we can find the value of xy (xy=n+2). Sufficient.
(2) x = y + 17 > xy=17. Directly gives us the value of xy. Sufficient.
Answer: D.
Hope it's clear.



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Re: Sammy has flavors of candies with which to make goody bags
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10 Jul 2018, 21:16
[quote="financestudent"]How do you make sure that it has only one solution without solving it, did we just assume it?
It may have 10 different roots. (let's say xy2 is a) C(a, 10) = ((a).(a1).(a2)....(a9))/10!=3003
Hello
You are right that we can simplify this as: ((a).(a1).(a2)....(a9))/10!=3003 We can say that ((a).(a1).(a2)....(a9)) = 3003*10!
Now since 3003*10! is a unique number (no matter how big it is), there can be ONLY single value of a positive integer 'a' such that ((a).(a1).(a2)....(a9)) equals 3003*10!. Since this is data sufficiency, we dont need to find the value of 'a'. If we know for sure that a unique value of 'a' exists, the statement is sufficient.



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Sammy has flavors of candies with which to make goody bags
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10 Jul 2018, 23:35
Hi there, thank you for the explanation. Can you also explain how you figured out that 3003*10! is a unique number or how can I tell that a number is unique (especially like this big one)? Bunuel amanvermagmatamanvermagmat wrote: financestudent wrote: How do you make sure that it has only one solution without solving it, did we just assume it?
It may have 10 different roots. (let's say xy2 is a) C(a, 10) = ((a).(a1).(a2)....(a9))/10!=3003
Hello
You are right that we can simplify this as: ((a).(a1).(a2)....(a9))/10!=3003 We can say that ((a).(a1).(a2)....(a9)) = 3003*10!
Now since 3003*10! is a unique number (no matter how big it is), there can be ONLY single value of a positive integer 'a' such that ((a).(a1).(a2)....(a9)) equals 3003*10!. Since this is data sufficiency, we dont need to find the value of 'a'. If we know for sure that a unique value of 'a' exists, the statement is sufficient.




Sammy has flavors of candies with which to make goody bags &nbs
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