Question 386910: Please help us understand this Algebra homework. I do not believe the teacher really explained this in class, and my child is confused. Thank you so much.
1) How many different license plates can be made that contain 4 letters and 3 numbers if repeating is allowed, but you cannot use Q or 7?
2) Using the digits 2,4,6,7,8 only?
a) How many 4-digit numbers can be made if repeating is allowed?
b) How many 4-digit numbers can be made if no repeating is allowed?
c) If you can repeat numbers, how many 3-digit numbers are possible that are greater than 600?
d) if no repeating is allowed, how many 5-digit numbers can you make that are greater than 60,000 and also odd?
Answer by sudhanshu_kmr(1152)   (Show Source):
You can put this solution on YOUR website!
1) License plates must be in the form LLLLNNN
where L represents a letter and N represents a number, here repetition is allowed but can't use Q or 7
so, we can use 25 letters and 9 numbers...
each letter place can be filled by any letter and each number place can be filled
by any number so,
no. of ways = 25*25*25*25*9*9*9
2) digits are 2,4,6,7,8
a) for 4 digit number ABCD,
any digit can be placed at each place
so, total no. of ways = 5*5*5*5 = 125
b) here repetition is not allowed
so, total no. of ways = 5* 4 * 3* 2 = 120
another method,
no. of ways = 5P4 = 120 (arrangement of 4 taking 5 digits)
c) here repetition is allowed to form 3 digit number i.e XYZ
at place of X we can put only 6, 7, or 8 i.e by 3 ways..
other places can be filled by any of 5 digits..
so, no. of ways = 3*5*5 = 75
d)
here repetition is not allowed to form 5 digit number i.e ABCDE
similar to previous problem, place A can be filled by only 3 ways..
but here 7 is only odd digit so, to form odd number 7 must be at place E.
thus place A can be filled by only 2 ways (either 6 or 8)
and now, places B, C AND D can be filled by any digit...
no. of ways to fill A = 2 (either 6 or 8)
no. of ways to fill B = 3 ( 2,4,6,8 except that on A)
no. of ways to fill C = 2 ( except on A and B)
no. of ways to fill D = 1 (only one will remain)
no. of ways to fill E = 1 (only 7)
total no. of ways = 2*3*2*1*1 = 12
so, total no. of odd numbers more than 60,000 = 12
if you also face difficulty to understand the concept, you are welcome to contact me by mail sudhanshu.cochin@gmail.com
|
|
|
