1. a restaurant offers a special pizza with any three toppings. if the restaurant has 15 toppings to choose…

1. a restaurant offers a special pizza with any three toppings. if the restaurant has 15 toppings to choose from, how many different special pizzas are possible?\n2. there are 20 skiers competing in the final round of the olympic freestyle skiing competition. in how many ways can three of the skiers finish first, second, and third to win the gold, silver, and bronze medals?\n3. a cooler has 8 bottles of coca - cola, 3 bottles of sprite, and 5 bottles of water. if you randomly select 4 bottles, what is the probability that you will select 2 bottles of coca - cola and 2 bottles of water?

1. a restaurant offers a special pizza with any three toppings. if the restaurant has 15 toppings to choose from, how many different special pizzas are possible?\n2. there are 20 skiers competing in the final round of the olympic freestyle skiing competition. in how many ways can three of the skiers finish first, second, and third to win the gold, silver, and bronze medals?\n3. a cooler has 8 bottles of coca - cola, 3 bottles of sprite, and 5 bottles of water. if you randomly select 4 bottles, what is the probability that you will select 2 bottles of coca - cola and 2 bottles of water?

Answer

Explicación:

Paso 1: Resolver el primer problema (combinaciones)

El número de combinaciones de $n$ elementos tomados $r$ a la vez se calcula con la fórmula $C(n,r)=\frac{n!}{r!(n - r)!}$. Aquí, $n = 15$ (número de toppings) y $r=3$ (número de toppings para la pizza). $$C(15,3)=\frac{15!}{3!(15 - 3)!}=\frac{15!}{3!×12!}=\frac{15\times14\times13\times12!}{3\times2\times1\times12!}=455$$

Paso 2: Resolver el segundo problema (permutaciones)

El número de permutaciones de $n$ elementos tomados $r$ a la vez se calcula con la fórmula $P(n,r)=\frac{n!}{(n - r)!}$. Aquí, $n = 20$ (número de esquiadores) y $r = 3$ (posiciones de medalla). $$P(20,3)=\frac{20!}{(20 - 3)!}=\frac{20!}{17!}=20\times19\times18 = 6840$$

Paso 3: Resolver el tercer problema (probabilidad)

  • En primer lugar, calculamos el número total de botellas: $8 + 3+5=16$ botellas.
  • Calculamos el número de maneras de elegir 4 botellas de 16, que es $C(16,4)=\frac{16!}{4!(16 - 4)!}=\frac{16\times15\times14\times13\times12!}{4\times3\times2\times1\times12!}=1820$.
  • Calculamos el número de maneras de elegir 2 botellas de Coca - Cola de 8: $C(8,2)=\frac{8!}{2!(8 - 2)!}=\frac{8\times7\times6!}{2\times1\times6!}=28$.
  • Calculamos el número de maneras de elegir 2 botellas de agua de 5: $C(5,2)=\frac{5!}{2!(5 - 2)!}=\frac{5\times4\times3!}{2\times1\times3!}=10$.
  • El número de maneras de elegir 2 botellas de Coca - Cola y 2 botellas de agua es $C(8,2)\times C(5,2)=28\times10 = 280$.
  • La probabilidad $P$ es $\frac{C(8,2)\times C(5,2)}{C(16,4)}=\frac{280}{1820}=\frac{2}{13}$.

Respuesta:

  1. 455
  2. 6840
  3. $\frac{2}{13}$