@ani6 said:PLZ help..From 6 couples who attended a party, four people are selected at random. Find the probability that exactly one couple is selected.
8/33?
@ani6 said:PLZ help..From 6 couples who attended a party, four people are selected at random. Find the probability that exactly one couple is selected.
6C1*10C1*8C1= 6*8*10
Prob = 6*8*10/ 12C4 ?
@anantn said:ah yes its 104.....my badhalf asleep, pardon my silly mistakes..
Check powers of 2 though....might be in for a surprise?
regards
scrabbler
@techgeek2050 @ScareCrow28 @ChirpiBird
regards
scrabbler
puys ans is 103
@scrabbler said:
Check powers of 2 though....might be in for a surprise?regardsscrabbler@techgeek2050@ScareCrow28@ChirpiBird
i gt u ...thanks
@anantn said:ah yes its 104.....my badhalf asleep, but too lazy to do college work, pardon my silly mistakes..
No it would be 103, as exponent of 2 in 10000! is 9995 and in 100! its 97
[9995/97] = 103
So, answer will be 103
Ideally one should check exponents of largest and lowest prime factors
@ani6 said:PLZ help..From 6 couples who attended a party, four people are selected at random. Find the probability that exactly one couple is selected.
is it 240/495 or 16/33 ??
{6c3*3c1 + 6c2*2c1*4c1 + 6c1*1c1*5c1}/12c4 = 240/495 = 16/33
@ChirpiBird said:can u plz post the solutionmy approach :number of zeroes in 10000! is 2499.. .. [this i found ...10000!/(5^n)] number of zeroes in 100! is 24so i can writeS*10^2499 = L*10^24*P24x = 2499x = 104
consider powers of 2..
@scrabbler said:Check powers of 2 though....might be in for a surprise?regardsscrabbler@techgeek2050@ScareCrow28@ChirpiBird
What is the fallacy??
@ScareCrow28 said:6C1*10C1*8C1= 6*8*10Prob = 6*8*10/ 12C4 ?
10C1 * 8C1 /2 since the order does not matter...So answer is 6 * 10 * 8/ 2 * 12C4 = 16/33 I guess.
regards
scrabbler
@Logrhythm said:is it 240/495 or 16/33 ??{6c3*3c1 + 6c2*2c1*4c1 + 6c1*1c1*5c1}/12c4 = 240/495 = 16/33
@scrabbler said:10C1 * 8C1 /2 since the order does not matter...So answer is 6 * 10 * 8/ 2 * 12C4 = 16/33 I guess.regardsscrabbler
And my answer (PnC) matches with that of Scrabbler.....let me take a moment and revel in that (glory) 
@scrabbler said:10C1 * 8C1 /2 since the order does not matter...So answer is 6 * 10 * 8/ 2 * 12C4 = 16/33 I guess.regardsscrabbler
Missed while typing@scrabbler said:10C1 * 8C1 /2 since the order does not matter...So answer is 6 * 10 * 8/ 2 * 12C4 = 16/33 I guess.regardsscrabbler
bro whats wrong with my approach
total selection of 4 people will be 12C4
Now to select a couple there can be 3M+1F or 3F+1M
So 3M or 3F can be selected in 6C3 and the remaining 1 M or F can be selected in 3C1 ways
prob=(6C3*3C1+6C3*3C1)/12C4=8/33
total selection of 4 people will be 12C4
Now to select a couple there can be 3M+1F or 3F+1M
So 3M or 3F can be selected in 6C3 and the remaining 1 M or F can be selected in 3C1 ways
prob=(6C3*3C1+6C3*3C1)/12C4=8/33
@iLoveTorres said:bro whats wrong with my approachtotal selection of 4 people will be 12C4Now to select a couple there can be 3M+1F or 3F+1MSo 3M or 3F can be selected in 6C3 and the remaining 1 M or F can be selected in 3C1 ways prob=(6C3*3C1+6C3*3C1)/12C4=8/33
u hv missed cases where there can be 1 couple and 2 female/males
and 6c3*3c1 would come only once, as the order is not necessary here...
6c3*3c1 --> 3 males/females and 1 of their spouse..
6c2*2c1*4c1 --> 2 males/females, 1 of their spouse and 1 male/female from any of the remaining 4 ppl...
6c1*1c1*5c1 --> 1 male/female, their spouse and 1 male/female from the remaining 5 ppl...
hope u got it...
@iLoveTorres said:bro whats wrong with my approachtotal selection of 4 people will be 12C4Now to select a couple there can be 3M+1F or 3F+1MSo 3M or 3F can be selected in 6C3 and the remaining 1 M or F can be selected in 3C1 ways prob=(6C3*3C1+6C3*3C1)/12C4=8/33
3M 1 F = 8/33.
Also 3F 1M = 8/33.
Total 16/33.
regards
scrabbler
Also 3F 1M = 8/33.
Total 16/33.
regards
scrabbler
@scrabbler said:3M 1 F = 8/33. Also 3F 1M = 8/33. Total 16/33.regardsscrabbler
i mean what's wrong in the logic
@ani6 said:@scrabbler10C1 * 8C1 /2???why divided by 2...plz elaborate??
Suppose the couples are Aa Bb Cc Dd Ee Ff
So first we choose a couple in 6 ways (let's say Ff)
Now if we choose the next person in 10 ways (suppose we choose D) then we can choose the other person in 8 ways (all but d). For example we could choose b. So 10 * 8. But.....
We counted the case of D followed by b. And we would also have a case of b followed by D which is the same but being counted as a separate one in the 10 * 8...
Hence double counting hence divide by 2.
(If not clear, think - if there are four people in a room and everyone shakes hands with everyone else, why are there 4*3/2 = 6 handshakes and not 4*3 = 12? Same funda)
regards
scrabbler
So first we choose a couple in 6 ways (let's say Ff)
Now if we choose the next person in 10 ways (suppose we choose D) then we can choose the other person in 8 ways (all but d). For example we could choose b. So 10 * 8. But.....
We counted the case of D followed by b. And we would also have a case of b followed by D which is the same but being counted as a separate one in the 10 * 8...
Hence double counting hence divide by 2.
(If not clear, think - if there are four people in a room and everyone shakes hands with everyone else, why are there 4*3/2 = 6 handshakes and not 4*3 = 12? Same funda)
regards
scrabbler