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 tablesSince, 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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