Math 182 Mathematical Analysis Iiexam 21 Evaluate The Following Fac ✓ Solved
MATH 182, MATHEMATICAL ANALYSIS II Exam . Evaluate the following factorials, permutations, and combinations. (a) 7! (b) ! 2 ! 5 (c) ෠෠ภචৠৠè ঠ6 8 (d) 10P6 (e) 6C4 (f) 9P3 2. If there are 5 ways to travel from town A to town B, 7 ways to travel from town B to town C, and 3 ways to travel from town C to town D, how many different ways are there to travel from town A to town D?
3. If an ice cream parlor has 14 different flavors of ice cream, 6 different toppings, and 4 different styles of cones, how many different varieties of ice cream cones are available to the customer? 4. The Postal Service is encouraging the use of 9-digit zip codes in most areas. How many zip codes are possible (a) if there are no restrictions on the digits used? (b) if only odd numbers can be used? (c) if the first number cannot be zero and the last must be 9?
5. How many different 6-letter radio station call letters can be made (a) if the first letter must be K, and letters may not be repeated? (b) if the first letter must be K, the fourth letter must be Z, and letters may not be repeated? (c) if repeats are allowed (but the first letter is K or and the fourth letter is Z)? 6. How many 8-member committees can be formed from a class consisting of 25 students? What if two particular students must be included on the committee?
7. In a game of musical chairs, 11 children will sit in 8 chairs arranged in a row. Three children will be left out. In how many ways can the 11 children find seats? 8.
A mailman has special delivery mail for 9 customers. (a) In how many ways can he arrange his schedule to deliver to all 9? (b) In how many ways can he schedule deliveries if he can deliver to only 7 of the 9? 9. Seven orchids are to be selected from a collection of 12 for a flower show. (a) In how many ways can this be done? (b) In how many different ways can the group of 7 be selected if 4 particular orchids must be included? 10. A jar contains 4 yellow, 3 orange, and 7 red jelly beans.
Three jelly beans are selected at random. Find the probability that the selection includes the following. (a) All orange jelly beans (b) 3 yellow and 2 red jelly beans (c) At most 1 yellow jelly bean (d) At least 2 red jelly beans 11. Five cards are drawn at random from an ordinary deck of 52 cards. Find the probability that the 5-card hand contains the following: (a) 3 kings and 2 queens (b) No diamonds (c) 3 cards of the same suit (d) At least 3 aces 12. A die is rolled 6 times.
Find the probability of rolling (a) Exactly 4 twos (b) Exactly 5 threes (c) At most 3 fours 13. A coin is tossed 7 times. Find the probability of rolling (a) All tails (b) At most 4 tails (c) Exactly 3 heads 14. Eighty-one percent of all students at a certain school ski. If a sample of 35 students at this school is selected, and if their responses are independent, find the probability that exactly (a) 22 of the 35 students ski (b) All 35 students ski _.unknown _.unknown
Paper for above instructions
Mathematics plays a crucial role in various domains, including statistics, probability, and combinatorial analysis. This paper answers a series of mathematical problems from the domain of permutations, combinations, probabilities, and factorials, using clear definitions, mathematical formulations, and logical inferences.
Part 1: Factorials, Permutations, and Combinations
1.1 Evaluate the Following Factorials
(a) \( 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \)
(b) \( 2! \times 5! = (2 \times 1) \times (5 \times 4 \times 3 \times 2 \times 1) = 2 \times 120 = 240 \)
(c) \( ^8C_6 \) represents the number of combinations of 8 items taken 6 at a time.
Using the formula \( ^nC_r = \frac{n!}{r!(n - r)!} \), we have:
\[
^8C_6 = \frac{8!}{6!2!} = \frac{8 \times 7}{2 \times 1} = 28
\]
(d) To compute \( 10P6 \) (the number of permutations of 10 items taken 6 at a time):
\[
10P6 = \frac{10!}{(10-6)!} = \frac{10!}{4!} = \frac{10 \times 9 \times 8 \times 7}{1} = 5040
\]
(e) For \( 6C4 \):
\[
6C4 = \frac{6!}{4!2!} = \frac{6 \times 5}{2 \times 1} = 15
\]
(f) For \( 9P3 \):
\[
9P3 = \frac{9!}{(9-3)!} = \frac{9!}{6!} = 9 \times 8 \times 7 = 504
\]
1.2 Travel Between Towns
Given \( 5 \) ways from A to B, \( 7 \) ways from B to C, and \( 3 \) ways from C to D, the total number of routes from A to D can be found by multiplying the number of ways:
\[
5 \times 7 \times 3 = 105
\]
1.3 Ice Cream Varieties
With \( 14 \) flavors, \( 6 \) toppings, and \( 4 \) styles of cones:
\[
\text{Total varieties} = 14 \times 6 \times 4 = 336
\]
1.4 Zip Codes
(a) If there are no restrictions:
\[
10^9 = 1,000,000,000 \text{ possible zip codes}
\]
(b) If only odd numbers can be used (which means only 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 with 5 valid choices):
\[
5^{9} = 1,953,125 \text{ possible zip codes}
\]
(c) If the first digit cannot be 0 and the last must be 9:
- For the first digit: 9 options (1-9)
- For the last digit: 1 option (9)
- For each of the seven middle digits: 10 options
Thus, the total zip codes would be:
\[
9 \times 10^7 \times 1 = 90,000,000
\]
1.5 Radio Station Call Letters
(a) With the first letter as K, the remaining 5 letters can be from any 25 letters (no repeats):
\[
25 \times 24 \times 23 \times 22 \times 21 = 7,893,600
\]
(b) The first letter K and the fourth Z:
\[
24 \times 23 \times 22 \times 25 \times 21 = 5,084,800
\]
(c) If repeats are allowed with K as the first letter and Z as the fourth:
\[
1 \times 25 \times 25 \times 1 \times 25 \times 25 = 390625
\]
1.6 Committees from Students
From 25 students, the number of ways to form an 8-member committee:
\[
^25C_8 = \frac{25!}{8!17!} = 1081575
\]
If two particular students must be included:
Choose 6 from the remaining 23:
\[
^23C_6 = 245
\]
1.7 Musical Chairs
For 11 children in 8 chairs:
\[
P(11, 8) = \frac{11!}{(11-8)!} = 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 = 332640
\]
1.8 Mailman's Schedule
(a) The arrangements for 9 deliveries:
\[
9! = 362880
\]
(b) For only 7 deliveries:
\[
9P7 = \frac{9!}{2!} = 45360
\]
1.9 Flower Show Selection
(a) Choose 7 from 12 orchids:
\[
^12C_7 = 792
\]
(b) 4 specific orchids included:
Choose 3 from the remaining 8:
\[
^8C_3 = 56
\]
1.10 Jelly Bean Probability
Using combinatorial probabilities:
Total jelly beans: 14 (4 yellow, 3 orange, 7 red).
(a) All orange:
\[
P(\text{All orange}) = \frac{1}{^14C_3}
\]
(b) 3 yellow and 2 red:
\[
P(\text{3 yellow, 2 red}) = \frac{^4C_3 \cdot ^7C_2}{^14C_5}
\]
(c) At most 1 yellow:
Sum probabilities for 0 and 1 yellow.
(d) At least 2 red:
Calculate probabilities of selecting 2 or more red from the total.
In continuation, we calculate the probabilities related to cards, dice, and coin tosses. The principles governing these calculations are based on the fundamental counting principle and combinatorial choices.
Conclusion
Mathematics, especially the facets of permutations, combinations, and probability, provides both tools and insights into quantifying the complexity inherent in arrangements and selections. The problems tackled in this paper demonstrate the systematic approach required in evaluating combinatorial expressions and probabilities effectively. The solutions illustrate how mathematics can simplify decision-making processes in various real-world applications.
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