permutations are permutations only

Haai brilliant members i am here to talk about how to find the no. of possible ways of arranging the letters of any word . i already gave a problem regarding this its too critical .

suppose there are 5 letters to be filled with vowels . we know that there are totally 5 vowels . in 1st blank there are 5 possibilities in which we can fill . in 2nd blank , we can fill in 4 possibilities because already we used a vowel to fill the 1st blank . same as 3rd with 3 vowels and 4th with 2 vowels and 5th with 1 vowel . so the total no. of ways is 5 X 4 X 3 X 2 X 1 = 5! = 120

there is also a formula for this it is n!/n - r where n stands for no. of letters and r stands for no. of blanks available to fill the letters . answer for the above example using formula is 5!/(5 - 5)! =120/1 = 120 [since 0! = 1] hope you understand see you next time like this for suggestions and doubts contact me at my e-mail "[email protected]"

#NumberTheory #Permutations #Notes

Easy Math Editor

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`2^{34}`	$2^{34}$
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Comments

vishal give a reply now

sudoku subbu - 6 years, 5 months ago

Please be mindful of your language on Brilliant. I have deleted your comment with offensive language.

Calvin Lin Staff - 6 years, 5 months ago

sorry for you . forgive me .

sudoku subbu - 6 years, 5 months ago

Nice.Thanks for you'r help.Even though I had known these already, It will be useful for me in the future

Vishal S - 6 years, 5 months ago

you are not upto my state : bose einstien condense state i want a note about bose einstien condense state and plasma state can you give it vishal if you are vishal give me that one

sudoku subbu - 6 years, 5 months ago

Okay. I'l give it in my note

Vishal S - 6 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$