Problem Solving

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 letter may be repeated and if the 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) 2951

B) 8125

C) 15600

D) 15302

E) 18278

abhi332 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 letter may be repeated and if the 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) 2951

B) 8125

C) 15600

D) 15302

E) 18278

For a one letter code there are 26 possibilities.

For a two letter code there are 26*26 possibilities

For a three letter code there are 26*26*26 possibilities

The total number of possibilities is 26 + 26*26 + 26*26*26 = 18278

Choose (e)

Of course, on the GMAT, if you find yourself cubing 26, you know there must be a quicker way to answer the question!

All powers of 6 end in 6. The units digit of 26 is 6. The units digits of both 26^2 and 26^3 will also be 6:

6+6+6 is 18. Therefore, the units digit of the final answer must be 8. Choose E.

Testluv wrote:Of course, on the GMAT, if you find yourself cubing 26, you know there must be a quicker way to answer the question!

All powers of 6 end in 6. The units digit of 26 is 6. The units digits of both 26^2 and 26^3 will also be 6:

6+6+6 is 18. Therefore, the units digit of the final answer must be 8. Choose E.

Good way to look at it. And a good point.

When I have to multiply large numbers I tend to factor. For example, I see 26^3 as

(25+1)(25+1)(25+1) = (675+1)(25+1) = 17576

That way works too!

Good way to look at it. And a good point.

When I have to multiply large numbers I tend to factor. For example, I see 26^3 as

(25+1)(25+1)(25+1) = (675+1)(25+1) = 17576

That way works too!

Yep, it sure does work. And, of course, the important thing is to come up with an approach that works for you on the spot.

That said, since the test is so strictly timed, it's always a good idea to review questions you've answered correctly, and to think about whether there is a quicker approach you could have employed to get to the correct answer. And, the more approaches we have in our toolkit on test day, the greater the liklihood that we will come up with an approach that works on the spot.

hi,

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!

abhi332 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 letter may be repeated and if the 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) 2951

B) 8125

C) 15600

D) 15302

E) 18278

I had a small question about this one

why can't we use 26C1 for one letter code = this of course will give 26
but 26C2 for two letter code is different from 26 *26
and 26c3 for 3 letter code is different from 26*26*26

26C2 doesn't this mean number of ways of selecting 2 letters from 26 letters ?
which cases are we missing when we do 26C2 .

so initially for this sum I went 26c1 * 26c2 *26c3

please help me understand why this approach is not suitable here. Thanks.

gmatter2012 wrote:

abhi332 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 letter may be repeated and if the 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) 2951

B) 8125

C) 15600

D) 15302

E) 18278

I had a small question about this one

why can't we use 26C1 for one letter code = this of course will give 26
but 26C2 for two letter code is different from 26 *26
and 26c3 for 3 letter code is different from 26*26*26

26C2 doesn't this mean number of ways of selecting 2 letters from 26 letters ?
which cases are we missing when we do 26C2 .

so initially for this sum I went 26c1 * 26c2 *26c3

please help me understand why this approach is not suitable here. Thanks.

Yes, I have the exact same question...Any takers?