numbers pb

[awax22]

two problems that I have a hard time understanding the answers:

1. How many 7-letter 'word' can be made using the letter C,C,D,D,D,A,E once each if the letter A and the letter E must not be separated by another letter ?
A)24; B)60; C)120; D)420; E)720

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

[camitava]

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.

[awax22]

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 ...

[gmatguy16]

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?

[gmatguy16]

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.

[awax22]

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

[Stuart@KaplanGMAT]

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.

[viidyasagar]

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

[sanju09]

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]