A)If a license plate consists of 2 letters followed by 3 digits or 3 letters followed by 2 digits, how many license plates can be created?

B)You are given an exam with only 9 questions. if each question has 4 choices, howmany different possible answers can you have for the entire exam?

C)If you have 20 students in your class, how many different ways can a committee of 5 be chosen if 1 person on the committee will be the president?

E) 25 people run a marathon. The first 6 people to finish will receive an award for their finish position. How many ways can the awards be given?

F) How many distinct arrangements of the letters in the word UNUSUAL can be made?

G) How many 4- letter words can be fromed by slection four letters from the word UNIFORM?

i know that you use ncr and ncp, but i don't understand can someone please show work! thanks for your time yoo

I assume the letter and numbers can be repeated, for example AA111

In that case all slots are independent, you do not have to do combinations
If there are 26 letters (in practice you do not want to use them all because 0 and O look to much alike etc)
ways to get 2 letters = 26*26
ways to get 3 digits = 10*10*10 (using 0--9)
so 26*26*1000
you can do the second half

http://www.jiskha.com/display.cgi?id=1232319439

http://www.jiskha.com/display.cgi?id=1232305810

B is done the same as A, 4^9 (huge)

C Ignoring the president bit, how many combinations of 20 students taken 5 at a time?
C(n,r) = n!/[(r!(n-r)! ]
= 20! / [5! (15!) ]
= 20*19*18*17 *16 /[5*4*3*2]
= 19 * 3* 17 * 16
now you have in each of these 4 groups of five students, five choices for president
so multiply that answer by 5*5*5*5

That is enough, you should be able to continue

E) 25 people run a marathon. The first 6 people to finish will receive an award for their finish position. How many ways can the awards be given?

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Well E is different from D in that the order of the 6 people matters (one gets blue ribbon, one red ribbon etc) so it is not just a combination of 6 out of 25
but is called the "permutations" of 25 taken 6 at a time:
P(n,r) = n!/(n-r)!
=25!/19! = 25*24 *....20

U N U S U A L

This would be permutations of 7 objects except that we have three identical Us.
so permutations of 7 objects = 7!
but we have embedded in there the permutations of three objects
3!
so I get
7!/3! = 7*6*5*4

thanks for your time and your help!

A) To calculate the number of license plates that can be created, we can break it down into two cases:

1. Two letters followed by three digits:
- For the first two letters, we have 26 options (26 letters in the alphabet).
- For the three digits, we have 10 options for each digit (0-9).
- Therefore, the number of plates that can be created in this case is: 26 * 26 * 10 * 10 * 10 = 676,000.

2. Three letters followed by two digits:
- For the three letters, we have 26 options for each letter.
- For the two digits, we have 10 options for each digit.
- Therefore, the number of plates that can be created in this case is: 26 * 26 * 26 * 10 * 10 = 17,576,000.

To calculate the total number of license plates that can be created, we sum the values from both cases: 676,000 + 17,576,000 = 18,252,000.

B) To calculate the number of different possible answers for the entire exam, we need to calculate the total number of possible combinations of answers for each question and then multiply it by the number of questions.

- Each question has 4 choices, so for each question, there are 4 possible answers.
- Since there are 9 questions in total, we multiply the number of choices for each question by itself 9 times (4^9).

Therefore, the total number of different possible answers for the entire exam is: 4^9 = 262,144.

C) To calculate the different ways a committee of 5 can be chosen from a class of 20 students, where 1 person is chosen as the president:

- We first need to choose the president from the 20 students, which can be done in 20 ways.
- After selecting the president, we need to choose the remaining 4 members of the committee out of the remaining 19 students. This can be calculated using the combination formula: C(n, r) = n! / (r!(n-r)!), where n is the total number of options and r is the number of choices we want to make.
- In this case, C(19, 4) = 19! / (4!(19-4)!) = 3876.

Therefore, the total number of different ways a committee of 5 can be chosen with a president from a class of 20 students is: 20 * 3876 = 77,520.

E) The number of ways the awards can be given depends on the number of finishers. Since the first 6 people will receive an award for their finish position, we have 6 choices for the first position, 5 choices for the second position, 4 choices for the third position, and so on, until 1 choice for the sixth position.

Therefore, the number of ways the awards can be given is: 6 * 5 * 4 * 3 * 2 * 1 = 720 ways.

F) To calculate the number of distinct arrangements of the letters in the word "UNUSUAL," we need to count the total number of letters in the word and consider that some letters may be repeated.

- The word "UNUSUAL" has 7 letters in total.
- Among these 7 letters, the letter "U" is repeated twice, so we need to account for that.

To calculate the number of distinct arrangements, we use the formula for permutations of a word with repeated letters: P(n; n1, n2, ..., nr) = n! / (n1! * n2! * ... * nr!), where n is the total number of letters and n1, n2, ..., nr represent the number of times each letter is repeated.

In this case, the number of distinct arrangements of the letters in the word "UNUSUAL" is: 7! / (2! * 2!) = 10,080.

G) To calculate the number of 4-letter words that can be formed by selecting four letters from the word "UNIFORM":

- The word "UNIFORM" has 7 letters.
- We want to select four letters to form a 4-letter word.

Using the combination formula, C(n, r) = n! / (r!(n-r)!), where n is the total number of options and r is the number of choices we want to make, we can calculate the number of 4-letter words.

In this case, C(7, 4) = 7! / (4!(7-4)!) = 35.

Therefore, the number of 4-letter words that can be formed by selecting four letters from the word "UNIFORM" is 35.