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There are x children at a birthday party, who will be seated at two di

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There are x children at a birthday party, who will be seated at two di [#permalink]

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New post 05 Jun 2015, 06:27
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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. \(\frac{(x-1)!}{(y-1)!(y-1)!}\)

B. \(\frac{x!}{y!(x-y)!}\)

C. \(\frac{y!}{x!(x-y)!}\)

D. \(\frac{(y-1)!}{(x-y)!(y-1)!}\)

E. \(\frac{(x-1)!}{(x-y)!(y-1)!}\)

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There are x children at a birthday party, who will be seated at two di [#permalink]

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New post 05 Jun 2015, 07:06
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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. \(\frac{(x-1)!}{(y-1)!(y-1)!}\)

B. \(\frac{x!}{y!(x-y)!}\)

C. \(\frac{y!}{x!(x-y)!}\)

D. \(\frac{(y-1)!}{(x-y)!(y-1)!}\)

E. \(\frac{(x-1)!}{(x-y)!(y-1)!}\)


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 tables

Since, It's already mentioned that Birthday girl Sally sits on the Cake table so now we are left with only (x-1) 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 (x-1) children to select from and (y-1) 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 = \(\frac{n!}{r!(n-r)!}\)

The ways of selecting (y-1) children out of (x-1) children for Cake table is given by (x-1)C(y-1) = \(\frac{(x-1)!}{(x-1-y+1)!(y-1)!}\)

i.e. \(\frac{(x-1)!}{(x-y)!(y-1)!}\)

Answer: Option
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Re: There are x children at a birthday party, who will be seated at two di [#permalink]

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New post 06 Jun 2015, 07:53
1
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. \(\frac{(x-1)!}{(y-1)!(y-1)!}\)

B. \(\frac{x!}{y!(x-y)!}\)

C. \(\frac{y!}{x!(x-y)!}\)

D. \(\frac{(y-1)!}{(x-y)!(y-1)!}\)

E. \(\frac{(x-1)!}{(x-y)!(y-1)!}\)



Since the Q speaks of group , so we are dealing with combination..
As sally is already part of one group, remaining x-1 have to be divided into two groups of y-1 and x-y(as y have alrady been fixed in other group)...
for choosing y-1 out 0f x-1= x-1 C y-1
=\(\frac{(x-1)!}{(y-1)!(x-1-(y-1))!}\)
=\(\frac{(x-1)!}{(y-1)!(x-y)!}\)
ans E
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Re: There are x children at a birthday party, who will be seated at two di [#permalink]

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New post 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. \(\frac{(x-1)!}{(y-1)!(y-1)!}\)

B. \(\frac{x!}{y!(x-y)!}\)

C. \(\frac{y!}{x!(x-y)!}\)

D. \(\frac{(y-1)!}{(x-y)!(y-1)!}\)

E. \(\frac{(x-1)!}{(x-y)!(y-1)!}\)


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.”
Image
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:
Image
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 \(\frac{7!}{4!3!}=\frac{(x-1)!}{(x-y)!(y-1)!}\).

The correct answer is E.

Attachment:
2015-06-08_1648.png
2015-06-08_1648.png [ 14.61 KiB | Viewed 1627 times ]

Attachment:
2015-06-08_1648_001.png
2015-06-08_1648_001.png [ 82.67 KiB | Viewed 1636 times ]

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Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: There are x children at a birthday party, who will be seated at two di [#permalink]

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New post 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.”
Attachments

birthday_1.PNG
birthday_1.PNG [ 103.58 KiB | Viewed 702 times ]


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Re: There are x children at a birthday party, who will be seated at two di   [#permalink] 11 Mar 2017, 12:36
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