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hree TAs are grading a final exam. There are a total of 60 exams to grade. (a) How many ways are there to distribute the exams among the TAs if all that matters is how many exams go to each TA? (b) Now suppose it matters which students' exams go to which TAs. How many ways are there to distribute the exams? (c) Suppose again that we are counting the ways to distribute exams to TAs and it matters which students' exams go to which TAs. The TAs grade at different rates, so the first TA will grade 25 exams, the second TA will grade 20 exams and the third TA will grade 15 exams. How many ways are there to distribute the exams?

Respuesta :

Answer:

a. 205320

b. 34220

c. 60! / (35)! (25)! + 60!/ (40)!(20)! + 60!/ (45)! (15)!

Step-by-step explanation:

a) The number of ways to dustribute exams among the TA's is:

n / (n - r)!

n= number of things to choose from

r= Choosing r number

60P3= 60! / (60 - 3)!

(60)(59)(58)(57)! / (57)!

=205320

B) The number of ways to dustribute the exams among the TA's is:

n! /(n - r)! r!

60C3= 60! /(60 - 3)! 3!

= 60!/ 57! 3!

= 60 × 59 × 58 / 3 × 2 × 1

= 34220

C) The required number of ways is:

60C25 + 60C20 + 60C15

= 60! / (35)! (25)! + 60!/ (40)!(20)! + 60!/ (45)! (15)!

MrRoyal

The TAs grading the final exam is an illustration of permutation and combination

  • If order matters, there are 205320 ways to distribute the exams
  • If order does not matter, there are 34220 ways to distribute the exams
  • There are ways [tex]\mathbf{1.69 \times 10^{26}}[/tex] to distribute at different rates of 25 exams, 20 exams and 15 exams

The given parameters are:

[tex]\mathbf{n = 60}[/tex] --- exam grades

[tex]\mathbf{r = 3}[/tex] --- number of TAs

(a) Distribute exams among the three TAs if order matters

This implies that we permute the 60 exams among the three TAs.

So, we have:

[tex]\mathbf{^{60}P_3 = \frac{60!}{(60 - 3)!}}[/tex]

This gives

[tex]\mathbf{^{60}P_3 = \frac{60!}{57!}}[/tex]

Expand

[tex]\mathbf{^{60}P_3 = \frac{60 \times 59 \times 58 \times 57!}{57!}}[/tex]

[tex]\mathbf{^{60}P_3 = 60 \times 59 \times 58}[/tex]

[tex]\mathbf{^{60}P_3 = 205320}[/tex]

(b) Distribute exams among the three TAs if order does not matters

This implies that we combine the 60 exams among the three TAs.

So, we have:

[tex]\mathbf{^{60}C_3 = \frac{60!}{(60 - 3)!}3!}[/tex]

This gives

[tex]\mathbf{^{60}C_3 = \frac{60!}{57!3!}}[/tex]

So, we have:

[tex]\mathbf{^{60}C_3 = \frac{205320}6}[/tex]

[tex]\mathbf{^{60}C_3 = 34220}[/tex]

(c) Distribute at different rates of 25 exams, 20 exams and 15 exams

This is calculated as:

[tex]\mathbf{Ways = \frac{60!}{25! \times 20! \times 15!}}[/tex]

So, we have:

[tex]\mathbf{Ways = 1.69 \times 10^{26}}[/tex]

Read more about permutation and combination at:

https://brainly.com/question/15301090

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