1
sanchit
·2009-12-29 04:35:10
(11!/5!6!)-(6!/5!).......................
13
Avik
·2009-12-29 04:45:44
Nopes sanchit....ans is in 2 digits only!
Thande dimaag se karne waala sawaal hai, Subjective type :P
1
taran
·2009-12-29 04:54:34
when A is the middle letter
then we have remaining 4 A's and 6 B's
now to have a palindrome
we shall have 2 A's and 3 A's before the middle letter and after that
(eg. AABBBABBBAA)
SO NO. OF WAYS = 5! / 2!3! = 10
B in middle is not possible
hw i counted one extra :P !!
13
Avik
·2009-12-29 05:03:39
@Tush... Iska kya matlab hai -"No .of ways in which A and B do not come together" ?
@taran... am still thinking over wat u did, will take some time :P , don't explain but let me get it...
Fr others-
Two imp. keypoints b4 starting -
1) Beginning & End letter should be the same.
2) But the Middle letter will always be "A" irrespective of whether the word begins with A or B !!
Then its easy....
13
Avik
·2009-12-29 22:36:03
Yeah avinav, answer's okay, but how did u make use of combinations?
n taran i got it, was getting confused with - "we shall have 2 A's and 3 A's .."-
instead of 2 A's n 3 B's.... small typo there...
1
Avinav Prakash
·2009-12-29 23:12:05
A has to be at the 6th posn in al cases.....to make the words symmetrical.....so on 1 side there r 5 digits.... 2 As and 3 bs..=5!3!2!=5C2=10
13
Avik
·2009-12-31 05:23:32
Yup, thts the crux...
I was wondering ki u did smthing like selecting n arranging etc. thts why asked...
So, finally the Qn has been defeated...[1]