11. There are eleven seniors and five juniors who are sprinters on the high school track team. The coach must select four sprint

Question

2 months ago

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11. There are eleven seniors and five juniors who are sprinters on the high school track team. The coach must select four sprint

ers to run the 800-meter relay race.
a. How many 4-sprinter relay teams can be formed from the group of 16 sprinters?
b. In how many ways can two seniors be chosen to be part of the relay team?
c. In how many ways can two juniors be chosen to be part of the relay team?
d. In how many ways can two seniors and two juniors be chosen to be part of the relay team?
e. What is the probability that two seniors and two juniors will be chosen for the relay team?

Mathematics

1 answer:

AnnZ [12.3K] · Answer 1 · 2026-03-22T06:12:02+03:00

Answer:

a) 1820

b) 55

c) 10

d) 550

e) 0.3022

Step-by-step explanation:

Data provided in the question:

Number of seniors = 11

Number of juniors = 5

n = total options

r = persons to select

number of combinations = $^nC_r$

a) The number of 4-sprinter relay teams from a total of 16 sprinters

n = 16

r = 4

= ¹⁶C₄

= $\frac{16!}{4!(16-4)!}$

= $\frac{16\times15\times14\times13\times12!}{4\times3\times2\times1(12!)}$

=

b) The ways to select two seniors for the relay team

n = 11

r = 2

= ¹¹C₂

= $\frac{11!}{2!(11-2)!}$

= $\frac{11\times10\times9!}{2\times1(9!)}$

=

c) The ways to choose two juniors for the relay team

n = 5

r = 2

= ⁵C₂

= $\frac{5!}{2!(5-2)!}$

= $\frac{5\times4\times3!}{2\times1(3!)}$

= 10

d) The combination of two seniors and two juniors for the relay team

= the number of ways to select two seniors × the number of ways to select two juniors

= 55 × 10

= 550

e) The probability that two seniors and two juniors are selected for the relay team

= [ the number of ways to select two seniors and juniors ] ÷ [the total number of possible 4-sprinter relay teams ]

= 550 ÷ 1820

= 0.30219 ≈ 0.3022