Question
How many 4-letter words can be formed by slection four letters from the word UNIFORM? I don't understand how to do this at all. I know the answer is 840 but why?
i have no idea how to simplify this... can anyone help?
(x+3)!/x+3
(x-2)!/x!
i have no idea how to simplify this... can anyone help?
(x+3)!/x+3
(x-2)!/x!
Answers
Reiny
Think of filling 4 compartments with different letters from UNIFORM
__*__*__*__*
you could fill the first place with any of the 7 letters, so there are 7 ways to do that...
<u>7</u>*__*__*__*
now that one letter is gone, you can fill the 2nd place with any of the 6 remaining letters ....
<u>7</u>*<u>6</u>*__*__*
that leaves any of the 5 remaining letters to go into the 3rd spot
<u>7</u>*<u>6</u>*<u>5</u>*__*
and finally any of the remaining 4 letters can go into the last place.
<u>7</u>*<u>6</u>*<u>5</u>*<u>4</u>*
7*6*5*4 = 840
You can use this type of reasoning for most of these questions.
__*__*__*__*
you could fill the first place with any of the 7 letters, so there are 7 ways to do that...
<u>7</u>*__*__*__*
now that one letter is gone, you can fill the 2nd place with any of the 6 remaining letters ....
<u>7</u>*<u>6</u>*__*__*
that leaves any of the 5 remaining letters to go into the 3rd spot
<u>7</u>*<u>6</u>*<u>5</u>*__*
and finally any of the remaining 4 letters can go into the last place.
<u>7</u>*<u>6</u>*<u>5</u>*<u>4</u>*
7*6*5*4 = 840
You can use this type of reasoning for most of these questions.
Reiny
for (x+3)!/x+3
let's illustrate with a numerical example
what is 5!/5
5! means 5*4*3*2*1
and when you divide that by 5, wouldn't that cancel the leading factor of 5 leaving you with
4*3*2*1 or 4!
in the same way
(x+3)! = (x+3)(x+2)(x+1)...(2)(1)
divide that by x+3 cancels the first factor leaving
(x+2)(x+1)...(2)(1) or (x+2)!
you do the last one, let me know what you got.
let's illustrate with a numerical example
what is 5!/5
5! means 5*4*3*2*1
and when you divide that by 5, wouldn't that cancel the leading factor of 5 leaving you with
4*3*2*1 or 4!
in the same way
(x+3)! = (x+3)(x+2)(x+1)...(2)(1)
divide that by x+3 cancels the first factor leaving
(x+2)(x+1)...(2)(1) or (x+2)!
you do the last one, let me know what you got.
Kennedy
i have no idea how to do the last one