euro wrote:A certain stock exchange designates each stock with a one-, two- or three-letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be repeated and if teh same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes?

(A) 2,951

(B) 8,125

(C) 15,600

(D) 16,302

(E) 18,278

OA is [spoiler](E)[/spoiler]

Total possible 1-letter codes = 26.
Total possible 2-letter codes = 26*26 = 676.
Total possible 3-letter codes = 26*26*26 = 17576

Total possible codes = 26 + 676 + 17576 = 18,278.

The correct answer is E.

@GmatGuryNY

I get confused on when to "add" combinations and when to "multiply" them.

For example with this question I thought the answer was 26^6 ... thus I looked for an answer with the last digit of 6 and could not find any.

What is the rule for when I am suppose to add combinations instead of multiply?

For example the classic question which asks you:

A single group of 5 people needs to be created consisting of 2 people from group A and 3 people from group B, group A has 6 people and group B has 5 people.

You figure out the combinations possible to pick 2 out of 6 (group a), and then MULTIPLY that number, by the number of combinations to pick 3 out of 5 (group b).

However in some cases you would need to add these combinations, instead of multiply, could you please clarify when that is?

Thanks!

zueswoods wrote:@GmatGuryNY

I get confused on when to "add" combinations and when to "multiply" them.

For example with this question I thought the answer was 26^6 ... thus I looked for an answer with the last digit of 6 and could not find any.

What is the rule for when I am suppose to add combinations instead of multiply?

For example the classic question which asks you:

A single group of 5 people needs to be created consisting of 2 people from group A and 3 people from group B, group A has 6 people and group B has 5 people.

You figure out the combinations possible to pick 2 out of 6 (group a), and then MULTIPLY that number, by the number of combinations to pick 3 out of 5 (group b).

However in some cases you would need to add these combinations, instead of multiply, could you please clarify when that is?

Thanks!

We can use the same reasoning that we would use for a probability problem:
When we want multiple options to happen together -- A AND B -- we MULTIPLY.
When we want one option to happen OR another option to happen -- A OR B -- we ADD.

A club consists of 5 men and 5 women. How many 4-member committees can be formed that include at least 3 women?

3 women and 1 man:
Number of ways to choose 3 women from 5 choices = 5C3 = (5*4*3)/(3*2*1) = 10.
Number of ways to choose 1 man from 5 choices = 5.
To form a committee with 3 women and 1 man, all of the options above must happen together.
Thus, we MULTIPLY:
10*5 = 50.

4 women:
Number of ways to choose 4 women from 5 choices = 5C4 = (5*4*3*2)/(4*3*2*1) = 5.

To form a committee with at least 3 women, we need either the first option (3 women and 1 man) OR the second option (4 women).
Thus, we ADD:
Total number of ways to form a committee with at least 3 women = 50+5 = 55.

One can speed up answering by just knowing that every possible scenario ends with 6. So 6x3 = _8

Dear Mitch
Tell me where I am wrong if i follow below procedure:
1) since 4 members are to be selected, I make 4 vacant position - - - -
2) At least 3 women are required, which can be satisfied by two combinations (3W and 1M) and all 4W.
3) First combination: W W W M
If I go with choices then it will be 5*4*3*5 = 300
4) Second combination: W W W W
If I go with choices then it will be 5*4*3*2 = 120
Hence total combinations = 300+120 = 420.

Thanks Guru!