MATH PLEASEEEEE
posted by Kennedy .
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 09)
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!(nr)! ]
= 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?

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!/(nr)!
=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!