Asked by avm
In how many of the arrangements would the two L's be together, for the word BASKETBALL?
Answers
Answered by
MathMate
The 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!/(...)
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!/(...)
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