For Q-1, IMO E.
For Q-2, 108 = 1 x 108
= 2 x 54
= 3 x 36
= 4 x 27
= 6 x 18
= 9 x 12
So total count = 12. IMO E.

For the question one, I don't see how it's even possible to get 720 possibilites.

The correct answer was 120, but I found 60 and I didn't know where I was wrong.

For the number 2, I got 12 as well, but the anwser was 10... Maybe they're wrong ...

for first question answer is 120...i solved it the traditional way and need someone to come up with a better solution..
first case i considered c and d together as groups and second case i considered both c's and d's as separate,there are 10 possible was to arrange them in 5 places(since a and e need to be together)
as groups: ccddd,dccdd,ddccd,dddccc
as separate: cdddc,cddcd,cdcdd,dcdcd,dcddc,ddcdc.
now a and e together have 6 places and they can be reveresed as ae and ea which gives 12 possibilities,hence answer is 12*10 = 120 ,amit if all the alphabets were distinct then number of possibilities =720.
for second question agree with 12,not sure why oa is 10.any inputs?

i am sorry amit ,if all alphabets are distinct then possibilities= 5 !*(6*2) =1440 assuming we have 5 distinct alphabets and a and e which need to be together.

Oh ok.. Thank you!!!

I understand now why i found 60... I didn't think of the switching of A and E. So I multiplied by 6 instead of 6x2

So the solution would be this:

6*2 (possibilities for A and E)*5! (ways of placing 5 letters) /(2!*3!)

we divide by 2! because there are two "C" which gives 2! permutation leading to the same word
and by 3! because there are 3 "D".

so it's 6*2*5!/(2!*3!)=12*10=120

To solve using the permutations formula, we need to treat A and E like one letter (since they always go together).

So, if we use the formula n!/r!s!, where r and s are the # of identical objects, we get:

6!/2!3! = 6*5*4*3!/2*3! = 6*5*4/2 = 120/2 = 60

Now, since we can have AE or EA, we need to double the number, so we end up with 60 * 2 = 120.

2. How many disctinct positive divisors does 108 have ?
A)6; B)9; C)10; D)11; E)12

108 = 2^2*3^3...Number of positive divisors (factors) = (2+1)*(3+1) = 3*4 = 12

Thumb rule = Write down the number as a product of its prime factors, i.e. N = a^p * b^q, then the number of positive factors is (p+1) * (q+1), answer is 12, choose E

Additionally, the sum of all those factors (if asked) is given by (a^(p+1) - 1)/ (a-1) * (b^(q+1) - 1)/ (b-1)

Hope this helps

2. How many disctinct positive divisors does 108 have ?
A)6; B)9; C)10; D)11; E)12

If X is a composite such that X = (a^m) (b^n) (c^p)..., where a, b, c,... are primes and m, n, p,... are positive integers; then the number of all distinct positive divisors of X is given by (m + 1)*(n + 1)*(p + 1)*...

Since, 108 = (2^2) (3^3), therefore the number of all distinct positive divisors of 108 will be given by (2 + 1)*(3 + 1) = [spoiler]12[/spoiler]. [spoiler]E[/spoiler]