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

[BTGmoderatorLU]

Beth's pizzeria offers x different toppings. What is the value of x?

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

The OA is B.

[deloitte247]

Question = What is x?
Statement 1 = There are an equal number of different pizzas that can be made with (x-2) toppings as these are different pizza's with just two toppings.
The number of x-2 topping pizzas that can be made out of x topping is;
$$C_x^2\ =\ \frac{x!}{\left(x-2\right)!2!}$$
The number of 2 topping pizzas that can be made out of
$$x\ toppings\ is\ C_x^2\ =\ \frac{x!}{\left(x-2\right)!2!}$$
$$The\ two\ \exp ressions\ for\ C_x^{x-2}\ and\ C_x^2\ are\ the\ same$$
So we cannot equate them and get the value of x
Hence, Statement 1 is INSUFFICIENT.

Statement 2 = If the pizzas were to add one additional topping, the number of different pizzas that could be made with 4 toppings would double.
This means that 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 4 x toppings is,
$$\ C_{x+1}^4\ =\frac{\left(6x+1\right)!}{4!\left(x-3\right)!}$$
The number of different pizzas that could be made with 4 toppings out of x toppings is,
$$\ C_{x+1}^4\ =\frac{x!}{4!\left(x-4\right)!}$$
$$\ Given\ that;\ \frac{\left(x\ +\ 1\right)!}{4!\left(x-3\right)!}=\ 2\ \left(\frac{x!}{4!\left(x-4\right)!}\right)$$
$$\ \frac{\left(x\ +\ 1\right)!}{\left(x-3\right)!}=\ 2\ \cdot\frac{x!}{\left(x-4\right)!}$$
$$\frac{\left(\ x!\ \cdot\ \left(x\ +\ 1\right)\right)}{\left(x-4\right)!\ \cdot\ \left(x-3\right)}=2\cdot\frac{x!}{\left(x-4\right)!}$$
$$\frac{\ \left(x\ +\ 1\right)}{\left(x-3\right)}=2$$
Cross multiply
x+1 = 2 * (x-3)
x + 1 = 2x - 6
2x - x = 1 + 6
x = 7

Statement 2 alone is SUFFICIENT.
Option B is CORRECT.

[fskilnik@GMATH]

Beth's pizzeria offers x different toppings. What is the value of x?

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

\[? = x\]
(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 2 toppings originally and 3 toppings when 1 topping is added, the number 0 is 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 thatr 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.