IIT JEE , AIEEE Forum - Mathematics , Algebra

heavensblaze

Find no. of ways of arranging the letters

AAAAA BBB CCC D EE F

in a row if the letters C are separated from one another.

btetarbe

You will have ve to use the complementary method for this.
All the possible permutations of the given set of alphabets
AAAAABBBCCCDEEF
is (15!)/(5!3!3!2!).
Now all the possible permutations with the 3 Cs together is(by considering the 3 Cs as 1 whole) (13!)/(5!3!2!).
Therefore, the possible arrangements of letters in a row keeping the Cs seperate
= No. of all possible arrangements - no. of possible permutations with the 3 Cs together
= (15!)/(5!3!3!2!) - (13!)/(5!3!2!)
Sorry I forgot abt the case of 2 Cs together. That makes the soln
= (15!)/(5!3!3!2!) - (13!)/(5!3!2!) - (14!)/(5!3!2!)

btetarbe

You will have ve to use the complementary method for this.
All the possible permutations of the given set of alphabets
AAAAABBBCCCDEEF
is (15!)/(5!3!3!2!).
Now all the possible permutations with the 3 Cs together is(by considering the 3 Cs as 1 whole) (13!)/(5!3!2!).
Therefore, the possible arrangements of letters in a row keeping the Cs seperate
= No. of all possible arrangements - no. of possible permutations with the 3 Cs together
= (15!)/(5!3!3!2!) - (13!)/(5!3!2!)

pannaguma

ans should be ( 13c3 * 12! )/5! * 3! * 2!.

yash_pandit

Sorry it was wrong

heavensblaze

Hey, pannaguma
How did you get this ans.
Plz explain

Asmita

PLZ C MY SOL. N TEMME IF DER'S NE FLAW IN DIS
total no. of possible permutations = 15!/5!3!3!2!
Total no. of possible permutations keeping d 3 c's together=13!/5!3!2!
total no. of possible permutations keeping d 2 c's together= 14!/5!3!2!
total no. of possible permutations keeping only 2 c's together at a tym= 14!/5!3!2! - 13!/5!3!2!
therefore
total no. of required permutations = 15!/5!3!3!2! - 13!/5!3!2! - (14!/5!3!2! - 13!/5!3!2!)
= 15!/5!3!3!2!- 14!/5!3!2!

heavensblaze

Hey Girls n boys ,
Thanks for your help.

I was able to work the problem out.

Just for your Knowledge the final answer is 95135040.

Thanks again

pannaguma

( 13c3 * 12! )/5! * 3! * 2!. = 95135040

priyesh

place the C's first

now four places are created

_ C _ C _ C _

now let x1 , x2 , x3 , x4 letters be placed in the gaps from L-R respectively

so since remaining letters are 12

hence

x1 + x2 + x3 + x4 = 12

since C's should be seperate x2 , x3 should not be zero

so applying this condition we find the no. of integral solns. for the above equation

no. of solns = 13C3

now these 12 letters can be arranged in 12!(5!*3!*2!)

hence no. of ways is

( 13C3 * 12! )/(5! * 3! * 2!).

pannaguma

priyesh i think you are beating around the bush in your first step.

just arrange the 12 non-C letters.
now put 3Cs in 13 places in 13C3 ways.

its the same thing but two ways of looking at it.

priyesh

oh yes u're right i've unneccesarily used a long method

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