In How many ways 8 distinct balls can be divided among 3 people such that anyone can get any number of balls? OA- 3^8Why cant we not perceive the problem as A+B+C=8where A,B,C are the number of balls each may receive. In that case number of solution is 8+3-1C2=10C2Please explain why cant we take this approach and where can we use this approach as well .@chillfactor@Subhashdec2@grkkrg@gs4890 PLEASE HELP !
a + b + c = 8 is used when the 8 items are identical.
Now one might argue that had the items been distinct then answer should be C(10, 2)*8!, but correct answer is 3^8.
Why ?
See, when we multiply C(10, 2) by 8!, then we are assuming that for all the cases considered in C(10, 2) we have 8! different way of doing that which is not correct.
For eg:- one of the case is when A gets 3, B gets 3 and C gets 2, this can be done in C(8, 3)*C(5, 3) = 8!/(3!3!2!)
So, its clear that we need to multiply by 8!/(3!3!2!) {and it will vary for all the remaining cases} and not 8!
3600?total number of arrangements = 8!number of ways in which all 3 vowels are together = 6! 3!number of ways in which any two vowels are together = 7! 2! 3number of ways in which vowels are not together = 8! - 6! 3! - 7! 2! 3 = 5760 ??
brdr it will use inclusion exclusion
in 7!2!3 u have all cases where 2 vowels are together and the rest of the alphabets are arranged so let itbe
(oe)sftwra
now when u arrange them there will be a case when (oe)axxxx this will be the arrangement
as u have minused them already in next step u need to add them
How many words can be made with word "SOFTWARE" in which vowels won't be together ?
_s_f_t_w_r_
6p3*5! = 14400 ?
@abhishek.2011 mera ans bhi 14400 hi aa raha tha....but everyone else were getting diff. ans..thats why didn't post my approach earlier.....apne upar confidence bikul zero hai...
_s_f_t_w_r_6p3*5! = 14400 ?@abhishek.2011 mera ans bhi 14400 hi aa raha tha....but everyone else were getting diff. ans..thats why didn't post my approach earlier.....apne upar confidence bikul zero hai...
arey yar post it galat hua to nobody s gonna bite u :P
brdr it will use inclusion exclusionin 7!2!3 u have all cases where 2 vowels are together and the rest of the alphabets are arranged so let itbe(oe)sftwranow when u arrange them there will be a case when (oe)axxxx this will be the arrangementas u have minused them already in next step u need to add themso 8! - 7! 2! 3 +3!6! = 14400
Thanx buddy, i got the flaw in my solution.
i have a doubt here. After subtracting 7! 2! 3 , we have actually subtracted all the cases wherein either 2 or 3 vowels appear together, right?
So adding 3!6! would mean we are adding the cases in which the 3 vowels are together, (which we have already subtracted), isn't it?
@vbhvguptaCannot be determined I think...take N as 0.000003logN = -6*log3 = -6*(0.477) = -2.862so greatest least integer becomes = - 2take N as 0.000009logN = -6*log9 = -6*2*log3 = -12*0.477 = -5.7 so the greatest integer becomes = -5Hence depends on the digit after the zeroes
N = .00000x = X*10^6 where x -> [1,10) logN = [-6 , -5 ) least integer for .000001 = -6 For all others = -5.____ CBD
Thanx buddy, i got the flaw in my solution.i have a doubt here. After subtracting 7! 2! 3 , we have actually subtracted all the cases wherein either 2 or 3 vowels appear together, right? So adding 3!6! would mean we are adding the cases in which the 3 vowels are together, (which we have already subtracted), isn't it?
ya but the problem lies with that 3c1
look u r subtracting them 2 times more so u have to add them once because it should be subtracted just 1nce :)
if still in doubt read inclusion exclusion principle from wikipedia.
