Math Problem Statement

valuate the expression.

7! − 5!   

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AUFEXC4 12.2.005.

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Evaluate the expression.

(6 − 3)!   

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AUFEXC4 12.2.015.

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Evaluate the expression.

P(8, 3) · P(4, 2)   

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AUFEXC4 12.2.047.

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A professor gives his students 9 essay questions to prepare for an exam. Only 4 of the questions will actually appear on the exam. How many different exams are possible?   different exams  

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AUFEXC4 12.2.021.

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Evaluate the expression.

C(19, 0)   

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AUFEXC4 12.2.027.MI.

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Evaluate the expression.   C(6, 4) · C(7, 5)

C(13, 9)      

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AUFEXC4 12.2.037.

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A student downloaded 5 music files to a portable music player. In how many different orders can the songs be played?

 different orders   

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AUFEXC4 12.2.049.

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A committee of 4 people is chosen from 9 women and 9 men. How many different committees are possible that consist of 2 women and 2 men?  committees  

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AUFEXC4 12.2.050.

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In a shipment of 21 smartphones, 2 are defective. How many ways can a quality control inspector randomly test 4 smartphones, of which 2 are defective?

 ways  

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AUFEXC4 12.2.065.

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This exercise refers to a standard deck of playing cards. Assume that 5 cards are randomly chosen from the deck.

How many hands contain 4 jacks?  hands  

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AUFEXC4 12.2.067.

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This exercise refers to a standard deck of playing cards. Assume that 5 cards are randomly chosen from the deck.

How many hands contain exactly 3 kings?  hands  

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AUFEXC4 12.2.070.

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This exercise refers to a standard deck of playing cards. Assume that 5 cards are randomly chosen from the deck.

How many hands contain exactly two 2s and two 5s?  hands  

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Solution

Let's evaluate each expression step by step.

1. 7!−5!7! - 5!

  • Factorial: n!n! means the product of all positive integers up to nn.
  • 7!=7×6×5×4×3×2×1=50407! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040
  • 5!=5×4×3×2×1=1205! = 5 \times 4 \times 3 \times 2 \times 1 = 120
  • Therefore, 7!−5!=5040−120=49207! - 5! = 5040 - 120 = 4920.

2. (6−3)!(6 - 3)!

  • 6−3=36 - 3 = 3, so we need to find 3!3!.
  • 3!=3×2×1=63! = 3 \times 2 \times 1 = 6.

3. P(8,3)×P(4,2)P(8, 3) \times P(4, 2)

  • Permutation: P(n,k)=n!(n−k)!P(n, k) = \frac{n!}{(n-k)!}.
  • P(8,3)=8!(8−3)!=8×7×61=336P(8, 3) = \frac{8!}{(8-3)!} = \frac{8 \times 7 \times 6}{1} = 336
  • P(4,2)=4!(4−2)!=4×31=12P(4, 2) = \frac{4!}{(4-2)!} = \frac{4 \times 3}{1} = 12
  • So, P(8,3)×P(4,2)=336×12=4032P(8, 3) \times P(4, 2) = 336 \times 12 = 4032.

4. Number of different exams from 9 questions where 4 will be chosen

  • Combination: C(n,k)=n!k!(n−k)!C(n, k) = \frac{n!}{k!(n-k)!}.
  • C(9,4)=9!4!(9−4)!=9×8×7×64×3×2×1=126C(9, 4) = \frac{9!}{4!(9-4)!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126.
  • There are 126 different possible exams.

5. C(19,0)C(19, 0)

  • Combination: C(19,0)=1C(19, 0) = 1 (Since choosing 0 items from a set is always 1 way).

Recap of Results:

  1. 7!−5!=49207! - 5! = 4920
  2. (6−3)!=6(6 - 3)! = 6
  3. P(8,3)×P(4,2)=4032P(8, 3) \times P(4, 2) = 4032
  4. 126 different exams
  5. C(19,0)=1C(19, 0) = 1

Would you like any more details on these calculations or have further questions?

Related Questions:

  1. How do you calculate P(10,4)P(10, 4)?
  2. What is the value of 8!8!?
  3. How many ways can you arrange 6 objects?
  4. What is C(12,5)C(12, 5)?
  5. How do you calculate C(8,2)×C(6,2)C(8, 2) \times C(6, 2)?

Tip:

Factorials grow very quickly, so when dealing with large numbers, it's essential to use a calculator or computational tools.

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Math Problem Analysis

Mathematical Concepts

Factorials
Permutations
Combinations

Formulas

n! (factorial)
P(n, k) (permutations)
C(n, k) (combinations)

Theorems

-

Suitable Grade Level

Advanced High School

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