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

Question

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)!}

GMATinsight · Accepted Answer

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  n C r  =  \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  E

chetan2u · Answer

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

Bunuel · Answer

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.” https://gmatclub.com/forum/download/file.php?id=26867 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: https://gmatclub.com/forum/download/file.php?id=26868 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. 2015-06-08_1648.png 2015-06-08_1648_001.png

anairamitch1804 · Answer

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.”

tarunanandani · Answer

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)!}

chetan2u · Answer

Please search before posting. You can ask your specific doubt in the thread

tarunanandani · Answer

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.

GyMrAT · Answer

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 =

(x-1)C(y-1) = \frac{(x-1)!}{(y-1)!*(x-y)!}

Answer E.

Thanks,
GyM

Villiavello · Answer

When reviewing the question, I tend to struggle to decipher a trap or if it is pertinent to the question. Example being:

"There are x children at a birthday party"  - wouldn't this include Sally, as she is a child who is at the party?

"exactly y children will be seated, including the birthday girl Sally".  - Since y children are seated there *including* the birthday girl, i.e. meaning that the birthday girl is already included in the number of "y".

I just wanted to ask if someone could explain to me with the example of this question, how to decipher whether a piece of information such as "including the birthday girl Sally" is a trap statement or should be included in the equation as it was in this answer.

bumpbot · Answer

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

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