Monday

September 1, 2014

September 1, 2014

Posted by **avm** on Wednesday, September 23, 2009 at 8:56pm.

- Gr. 12 Data Management -
**MathMate**, Wednesday, September 23, 2009 at 10:48pmThe number of arrangements for n distinct lettes is

n!

For example: 4!=24 distinct words can be made from the letters ABCD.

The number of arrangements for n letters, of which p are identical is

n!/p!

For example: 4!/2!=12 words can be made from the letters AABC.

The number of arrangements for n letters, of which p are identical and q are identical is

n!/(p!q!)

For example: 6!/(2!2!)=180 words can be made from the letters AABBCD.

If two letters have to be together all the timee, treat them as a single letter.

For the word BASKETBALL,

first arrange them in alphabetical order:

AABBEKLLST

Out of the 10 letters, AA and BB are repetitions, LL can be treated as one single letter (to have a total of 11 letters).

So the number of distinct words possible is

11!/(...)

**Related Questions**

Data Management - At a banquet, 4 coupes are sitting along one side of a table ...

Gr. 12 Data Management - How many arrangements of the word ALGORITHM begin with ...

Management 330 - Please can you help me put an Organizing Paper together: 1050 ...

English - 1. Put two scrambled word pieces together and make one complete word. ...

Gr. 12 Data MAnagement - In how many ways can the letter of the word SECTION be ...

math 12 - Find the number of seating arrangements of eight basketball players on...

Design - What is the difference between the vegetative style and the landscape ...

Math - How many ways can seven children sit on a bench? How many of these ...

Transition Math - Can you help discribe how to do this or what it will look like...

Education/Technology - The type of software that uses fields, records, and files...