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There are x children at a birthday party, who will be seated at two di
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05 Jun 2015, 06:27
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There are x children at a birthday party, who will be seated at two di
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05 Jun 2015, 07:06
Bunuel wrote: There are x children at a birthday party, who will be seated at two different tables. At the table with the birthday cake on it, exactly y children will be seated, including the birthday girl Sally. How many different groups of children may be seated at the birthday cake table?
A. \(\frac{(x1)!}{(y1)!(y1)!}\)
B. \(\frac{x!}{y!(xy)!}\)
C. \(\frac{y!}{x!(xy)!}\)
D. \(\frac{(y1)!}{(xy)!(y1)!}\)
E. \(\frac{(x1)!}{(xy)!(y1)!}\) The question mentioned that "How many different groups of children may be seated at the birthday cake table?" Therefore, we must understand that this doesn;t include any arrangement of groups on the table and this deals with only pure selection of groups of children to be seated on two different tablesSince, It's already mentioned that Birthday girl Sally sits on the Cake table so now we are left with only (x1) children to select the group of children to be seated on two tables Also, we must understand that one group selected will cause another group to form automatically so we only have to select the group for one table Let's select the group for the Cake table We have (x1) children to select from and (y1) children are to be selected for Cake table as one of the Child has to be the Birthday Girl The ways to select n out of r is given by nCr = \(\frac{n!}{r!(nr)!}\)The ways of selecting (y1) children out of (x1) children for Cake table is given by (x1)C(y1) = \(\frac{(x1)!}{(x1y+1)!(y1)!}\) i.e. \(\frac{(x1)!}{(xy)!(y1)!}\) Answer: Option
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Re: There are x children at a birthday party, who will be seated at two di
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06 Jun 2015, 07:53
Bunuel wrote: There are x children at a birthday party, who will be seated at two different tables. At the table with the birthday cake on it, exactly y children will be seated, including the birthday girl Sally. How many different groups of children may be seated at the birthday cake table?
A. \(\frac{(x1)!}{(y1)!(y1)!}\)
B. \(\frac{x!}{y!(xy)!}\)
C. \(\frac{y!}{x!(xy)!}\)
D. \(\frac{(y1)!}{(xy)!(y1)!}\)
E. \(\frac{(x1)!}{(xy)!(y1)!}\) Since the Q speaks of group , so we are dealing with combination.. As sally is already part of one group, remaining x1 have to be divided into two groups of y1 and xy(as y have alrady been fixed in other group)... for choosing y1 out 0f x1= x1 C y1 =\(\frac{(x1)!}{(y1)!(x1(y1))!}\) =\(\frac{(x1)!}{(y1)!(xy)!}\) ans E
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Re: There are x children at a birthday party, who will be seated at two di
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08 Jun 2015, 05:53
Bunuel wrote: There are x children at a birthday party, who will be seated at two different tables. At the table with the birthday cake on it, exactly y children will be seated, including the birthday girl Sally. How many different groups of children may be seated at the birthday cake table?
A. \(\frac{(x1)!}{(y1)!(y1)!}\)
B. \(\frac{x!}{y!(xy)!}\)
C. \(\frac{y!}{x!(xy)!}\)
D. \(\frac{(y1)!}{(xy)!(y1)!}\)
E. \(\frac{(x1)!}{(xy)!(y1)!}\) MANHATTAN GMAT OFFICIAL SOLUTION:This is a VICs problem, so we can pick numbers and test the answer choices: x = 8 y = 5 There are 8 children at the party, and 5 will sit at the table with the cake. Sally must sit at the birthday cake table, so we must pick 5 – 1 = 4 of the other 8 – 1 = 7 children to sit at that table with her. How many different ways can we choose 4 from a group of 7? Let's set up an anagram grid, where Y means “at the cake table” and N means “at the other table.” Now we can calculate the number of possible groups: \(\frac{7!}{4!3!}=35\) Our target answer is 35. Now we can test each answer choice by plugging in x = 8 and y = 5: Only choice E results in the target number.As an alternative to testing all five choices, we could have set up a formula using our selected numbers, then stopped to think about where those numbers originated. The number of possible groups was \(\frac{7!}{4!3!}=35\), but remember that this formula took Sally into account. The 7 came from 8 – 1 = x – 1. The 4 came from 5 – 1 = y – 1. The 3 came from the difference between these numbers: 7 – 4 = (x – 1) – (y – 1) = (x – y). Substituting these variable expressions in place of the numbers, we get \(\frac{7!}{4!3!}=\frac{(x1)!}{(xy)!(y1)!}\). The correct answer is E.Attachment:
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Re: There are x children at a birthday party, who will be seated at two di
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11 Mar 2017, 12:36
This is a VICs problem, so we can pick numbers and test the answer choices: x = 8 y = 5 There are 8 children at the party, and 5 will sit at the table with the cake. Sally must sit at the birthday cake table, so we must pick 5 – 1 = 4 of the other 8 – 1 = 7 children to sit at that table with her. How many different ways can we choose 4 from a group of 7? Let's set up an anagram grid, where Y means “at the cake table” and N means “at the other table.”
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There are x children at a birthday party, who will be seated
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28 Jun 2018, 06:13
There are \(x\) children at a birthday party, who will be seated at two different tables. At the table with the birthday cake on it, exactly \(y\) children will be seated, including the birthday girl Sally. How many different groups of children may be seated at the birthday cake table? A. \(\frac{(x1)!}{(y1)!(y1)!}\) B. \(\frac{x!}{y!(xy)!}\) C. \(\frac{y!}{x!(xy)!}\) D. \(\frac{(y1)!}{(xy)!(y1)!}\) E. \(\frac{(x1)!}{(xy)!(y1)!}\)
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Re: There are x children at a birthday party, who will be seated at two di
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28 Jun 2018, 06:28
tarunanandani wrote: There are \(x\) children at a birthday party, who will be seated at two different tables. At the table with the birthday cake on it, exactly \(y\) children will be seated, including the birthday girl Sally. How many different groups of children may be seated at the birthday cake table?
A. \(\frac{(x1)!}{(y1)!(y1)!}\)
B. \(\frac{x!}{y!(xy)!}\)
C. \(\frac{y!}{x!(xy)!}\)
D. \(\frac{(y1)!}{(xy)!(y1)!}\)
E. \(\frac{(x1)!}{(xy)!(y1)!}\) Please search before posting. You can ask your specific doubt in the thread
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1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
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Re: There are x children at a birthday party, who will be seated at two di
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28 Jun 2018, 06:46
chetan2u wrote:
Please search before posting. You can ask your specific doubt in the thread
I was able to solve the problem by assuming values for x and y but it took me more than 2.5 mins to solve it, so I was looking for another way of solving it. And I was able to find that in this thread. Thanks for the help. P.S. I'm sorry, I tried to search the question in the forum but couldn't find it. Will be more careful from next time.



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Re: There are x children at a birthday party, who will be seated at two di
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28 Jun 2018, 10:40
Bunuel wrote: There are x children at a birthday party, who will be seated at two different tables. At the table with the birthday cake on it, exactly y children will be seated, including the birthday girl Sally. How many different groups of children may be seated at the birthday cake table?
A. \(\frac{(x1)!}{(y1)!(y1)!}\)
B. \(\frac{x!}{y!(xy)!}\)
C. \(\frac{y!}{x!(xy)!}\)
D. \(\frac{(y1)!}{(xy)!(y1)!}\)
E. \(\frac{(x1)!}{(xy)!(y1)!}\) Total children at the birthday party including the birthday girl "\(x\)" Total children excluding the birthday girl \((x  1)\) # of children to be selected to sit at the birthday cake table, excluding the birthday girl \((y  1)\). Hence we need to select \((y  1)\) children out of \((x  1)\) children to sit with the birthday girl at the birthday cake table = \((x1)C(y1) = \frac{(x1)!}{(y1)!*(xy)!}\) Answer E. Thanks, GyM
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