Beth's pizzeria offers x different toppings. What is the val

Bunuel, could you please elaborate the second statement ? Especially how to expand this (X-3)! And (X-4)! ?

Thanks.

(2) If the pizzeria were to add one additional topping, the number of different pizzas that could be made with 4 toppings would double.

So, we are told that the number of different pizzas that could be made with 4 toppings out of x+1 toppings is twice the number of different pizzas that could be made with 4 toppings out of x toppings.

The number of different pizzas that could be made with 4 toppings out of x+1 = \(C^4_{x+1}=\frac{(x+1)!}{4!(x-3)!}\).

The number of different pizzas that could be made with 4 toppings out of x toppings = \(C^4_x=\frac{x!}{4!(x-4)!}\).


Thus given that \(\frac{(x+1)!}{4!(x-3)!}=2*\frac{x!}{4!(x-4)!}\);

\(\frac{(x+1)!}{(x-3)!}=2*\frac{x!}{(x-4)!}\);

Since \((x+1)! = x!*(x+1)\) and \((x-3)! = (x-4)!*(x-3)\), then \(\frac{ x!*(x+1)}{(x-4)!*(x-3)}=2*\frac{x!}{(x-4)!}\);

\(\frac{(x+1)}{(x-3)}=2\);

\(x+1=2*(x-3)\) --> \(x=7\).

Hope it's clear.

Bunuel, perfect !

Thank you.

Hi All,

This question suffers a bit from "interpretative bias", which is something that you're not likely to see on Test Day. While Bunuel's explanation is spot-on, the math involved is not obvious from the wording of the question. Here's why:

From Fact 1....

(1) There are an equal number of different pizzas that can be made with (x − 2) toppings as there are different pizzas with just 2 toppings.

The phrase "different pizzas with just 2 toppings" COULD mean....

1) There are 2 toppings to choose from. You can have a pizza with 0 toppings, 1 topping or 2 toppings.

OR

2) It's a reference to ALL of the different pizzas that COULD be made that have 2 toppings on them (from the X toppings that are there).

These statements involve different math and end in different results (which obviously impacts what the correct answer will be). Thankfully, the GMAT is really rigorous with its design standards and has an active process to weed out these type of inconsistencies, so you're not likely to face this issue on the Official GMAT.

GMAT assassins aren't born, they're made,
Rich

(1) There are an equal number of different pizzas that can be made with (x − 2) toppings as there are different pizzas with just 2 toppings.

As we know that nCr = nCn-r , this statement gives no info. hence insufficient
(2) If the pizzeria were to add one additional topping, the number of different pizzas that could be made with 4 toppings would double.

Here (x+1)C 4 = 2 * x C4, It can be solved for x. Hence sufficient.

\(? = x\,\,\)

(1) In this statement we must implicitly assume that x is at least 3, because no pizzas can be made with x-2 toppings if x is less than that.

> If there are exactly x=3 toppings, with 3-2=1 topping we can make 3 pizzas, exactly the same number of different pizzas we can make with just 2 toppings: C(3,2) = 3.
> If there are exactly x=4 toppings, with 4-2=2 toppings we can make exactly the same number of different pizzas that with just 2 toppings... sure!


(2) In this statement we must implicitly assume that x is at least 4 (*), because no pizzas could be made with 4 toppings if x is less than that.

(If, for instance, we had only x+1=3 toppings after 1 topping was added, the number 0 would be equal to twice itself, but we are not dealing with this scenario as viable.)

The number of different pizzas that can be made with 4 toppings among x+1 toppings (x at least 4) is

\(C\left( {x + 1,4} \right)\,\, = \,\,\frac{{\left( {x + 1} \right)!}}{{\,4!\,\,\left( {x - 3} \right)!\,}}\,\,\, = \,\,\,\frac{{\left( {x + 1} \right)x\left( {x - 1} \right)\left( {x - 2} \right)}}{{4!}}\)

and the number of different pizzas that can be made with 4 toppings among x toppings (x at least 4) is

\(C\left( {x,4} \right) = \frac{{x!}}{{\,4!\,\,\left( {x - 4} \right)!\,}} = \frac{{x\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}{{4!}}\)

Therefore, using statement (2), we know that

\(\frac{{\left( {x + 1} \right)x\left( {x - 1} \right)\left( {x - 2} \right)}}{{4!}} = 2 \cdot \frac{{x\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)}}{{4!}}\,\,\,\,\,\,\mathop \Rightarrow \limits^{:\,\,\frac{{x\left( {x - 1} \right)\left( {x - 2} \right)}}{{4!}}\,\, \ne \,0\,\,\,\left( * \right)} \,\,\,\,\left( {x + 1} \right) = 2\left( {x - 3} \right)\)

\(\left( {x + 1} \right) = 2\left( {x - 3} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,x = 7\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{SUFF}}.\,\,\,\,\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

(1) There are an equal number of different pizzas that can be made with (x − 2) toppings as there are different pizzas with just 2 toppings.

Through the combination we should understand the fact that XC2 and XCx-2 should be the same since they both yields same factorial value


(2) If the pizzeria were to add one additional topping, the number of different pizzas that could be made with 4 toppings would double.

Through this we can extract the exact value since X+1C4 =2*XC4

We can get x=7

Hence IMO B

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