Combinatorics Problem Set

[bangu]

Source: Word Translations GMAT Prep Guide
Location: Combinatorics Problem Set (Question Number 3)

Solution says that each number can be Ten's digits or Unit's digits or not a digit in the number. So possible cases are
5!/2!*3! = 10.

But what about substracting 5 numbers from it, which are double number 11, 22, 33, 44, and 55 (as mentioned in the question that digits should not be repeated).

Correct me if I am wrong.

[Guest]

Source: Word Translations GMAT Prep Guide
Location: Combinatorics Problem Set (Question Number 3)

Solution says that each number can be Ten's digits or Unit's digits or not a digit in the number. So possible cases are
5!/2!*3! = 10.

But what about substracting 5 numbers from it, which are double number 11, 22, 33, 44, and 55 (as mentioned in the question that digits should not be repeated).

Correct me if I am wrong.

Ignore earlier message, instead it should read like this

Location: Combinatorics Problem Set (Question Number 3)

Solution says that each number can be Ten's digits or Unit's digits or not a digit in the number. So possible cases are
5!/3! = 20.

But what about substracting 5 numbers from it, which are double number 11, 22, 33, 44, and 55 (as mentioned in the question that digits should not be repeated).

Correct me if I am wrong.

[Spencer]

Source: Word Translations GMAT Prep Guide
Location: Combinatorics Problem Set (Question Number 3)

Solution says that each number can be Ten's digits or Unit's digits or not a digit in the number. So possible cases are
5!/2!*3! = 10.

But what about subtracting 5 numbers from it, which are double number 11, 22, 33, 44, and 55 (as mentioned in the question that digits should not be repeated).

Correct me if I am wrong.

Ignore earlier message, instead it should read like this

Location: Combinatorics Problem Set (Question Number 3)

Solution says that each number can be Ten's digits or Unit's digits or not a digit in the number. So possible cases are
5!/3! = 20.

But what about subtracting 5 numbers from it, which are double number 11, 22, 33, 44, and 55 (as mentioned in the question that digits should not be repeated).

Correct me if I am wrong.

The question is:

A bball league assigns every player a two digit number for the back of a jersey from 1-5. What is the max. number of players that can join the league if no player has a number with a repeated digit (Ex: 22). and no two players have the same #.

Here's an easier way to understand it.

This problem is telling us that we have 5 different things and wants to know how many different groups of 2 can be made. So for example, since 1 and 2 on a jersey are different then 2 and 1 ORDER DOES MATTER.

What if the question was, "How many different ways can Bill, Joe, Mike, Steve, and Rick be grouped in two's?"
Would you find the answer and subtract 5 for the combo's of Bill and Bill, Mike and Mike...etc.? No sir.

Our calculations won't include the doubled up numbers because that's not how the combination formula works and its simply put in there so you don't add them in at the end.

The MGMAT gives you an anagram method to figure out these problems. Here's an easy way to understand

1) Make a simple chart with empty boxes. The number of boxes equals the items you're working with

2) Assign different variables for the "open spots" on your chart and fill in the rest of the spaces on the chart with X's.


[/img][/u][img]http://img143.imageshack.us/img143/9950/charts3kf0.jpg


3) Setup a simple division equation and take the total number of spaces in your chart and put it on top of the equation followed by a "!"

4) Add up the number of X's and put it on the bottom followed by a "!"

5! / 3! = 20

I usually go old school analog for these kinds of problems because it can be solved quickly and checking your answer is super easy.

1 : 2, 3, 4, 5 (4 players/numbers)
2 : 1, 3, 4, 5 (4 players/numbers)
3 : 1, 2, 4, 5 (4 players/numbers)
4 : 1, 2, 3, 5 (4 players/numbers)
5 : 1, 2, 3, 4 (4 players/numbers)

4 players x 5 options = 20[/img]

[rfernandez]

Bangu,

You used the permutation formula: P(n,r) = n!/(n-r)!. And this formula assumes that once an element is picked (in this case, a digit between 1 and 5) it can't be picked again. That's why you get 20 right away with the formula and don't have to subtract the 5 double-digit numbers.

Another approach is to say that you have 5 choices for the tens digit and 5 choices for the units digit, giving 5 * 5 = 25 total options. In this case, it is possible to pick repeated digits, so you'd subtract the 5 double-digit numbers from 25, to give you 20.

Rey