A tough math question, including permutations?

Question

Ok here it goes...

suppose certain account numbers are to consist of two letters using letters from the set L={a,b,c,d} followed by three digits from the set N={1,2,3,4,5} and then three more letters from L, where repetitions of letters or digits are not allowed within any of the three groups, but the last group of letters may contain one or both of those used in the first group. How many such account numbers are possible?

thanks in advance for sombodys help, anybodys help...

Answer 1

There are 4*3*5*4*3*4*3*2=17280 possible account numbers.

Answer 2

Possibilities of having two letters using letters from the set L are 4*3=12
Possibilities of having three digits using No's from the set N are 5*4*3=60
B'coz the last group of letters may contain one or both of those used in the first group now
Possibilities of having three letters using letters from the set L are
4*3*2=24
Now the total no of possibilities to have such an account no will 12*60*24

Answer 3

first group:
There are are 4*3=12 possibilities of having two letters using letters from the set L.

second group:
There are are 5*4*3=60 possibilities of having three digits using No's from the set N.

third group:

There are 4*3*2=24 possibilities of having three letters using letters from the set L.
B'coz the last group of letters may contain one or both of those used in the first group now

Therefore the total no of possibilities to have such an account no will be equal to the product of the three group possibilities.

12*60*24 = 17,280

I want 10 points plssssssssssssss!

Answer 4

Hi,
You can do this in two ways. (the second way is a bit easier to read..you might want to start with that one)
First, you can use the formula n!/(n-r)!, to find the number of ordered subsets of size r out of an n-element set. So for your first group, you want the number of ordered subsets of size 2 from a 4-element set, so (4!/2!) ways two pick the first two letters. Then there are (5!/(5-3)!)=(5!/2!) ways to pick the three numbers. And then, (4!/1!) ways to pick the last three letters. So, we have (4!/2!)((5!/2!)(4!/1!) account numbers possible.

You can also do it this way. You have (4 choose 2) ways to choose 2 elements out of a four element set and then 2! ways to order them, so you have (4 choose 2)(2!) ways to pick the first two letters. Then (5 choose 3) ways to choose 3 numbers out of five and 3! ways to order those numbers. And (4 choose 3) ways to choose 3 letters out of four and 3! ways to order them. Hence, you have
(4 choose 2)(2!)(5 choose 3)(3!)(4 choose 3)(3!) account numbers.
Hope that helps!

Answer 5

(4*3)*(5*4*3)*(4*3*2)=17280
got the point?
1st set 4*3 is coz u shud select 2 different letters from 4 letters.
1st letter has 4 choices n 2nd has 3....coz already 1 is selected n u hav only 3 more left
same logic applies 4 da rest..
its not a tough one indeed...its from very basics of permutations...

Answer 6

N = 4*3*5*4*3*4*3*2
N = 12*60*24
N = 288*60
N = 17,280