"Two solutions having milk and water in the ratios of 3:5 and 4:7 are mixed in the ratio of 2:3. Find the ratio of milk and water in the resultant mixture."
@pirateiim478 said:Rem (16!+91/323) =Rem(16!+91/17*19)= Rem((16!/17)*1/19+(91/17)*19)Rem((16+6)/19) =3
Rem(16!+91/17*19)= Rem((16!/17)*1/19+(91/17)*19) is wrong mate !!
@veertamizhan said:"Two solutions having milk and water in the ratios of 3:5 and 4:7 are mixed in the ratio of 2:3. Find the ratio of milk and water in the resultant mixture."
required ratio= (3/8)*2+(4/11)*3:(2+3)
=81:220
A die is thrown and a coin is tossed as many number of times as the number on the die. What is the probability of getting more heads than tails ?
@veertamizhan said:"Two solutions having milk and water in the ratios of 3:5 and 4:7 are mixed in the ratio of 2:3. Find the ratio of milk and water in the resultant mixture."
M..................W
3/8*2/5..........5/8*2/5
4/11*3/5........7/11*3/5
(3/8*2/5+4/11*3/5)/(5/8*2/5+7/11*3/5)=81/139?
In how many ways can 4 balls of different colors put in 3 urns which are colored black?
@bs0409 said:In how many ways can 4 balls of different colors put in 3 urns which are colored black?
3^4 ??
@Budokai001 said:A die is thrown and a coin is tossed as many number of times as the number on the die. What is the probability of getting more heads than tails ?
Let X=result on DIE
Now P(H>T|X=1)=1/2
Now P(H>T|X=2)=1/4
Now P(H>T|X=3)=1/2
Now P(H>T|X=4)=5/16
Now P(H>T|X=5)=1/2
Now P(H>T|X=6)=22/64
So P(H>T)=(1/6)*(1/2+1/4+1/2+5/16+1/2+22/64)=77/192
=
@bs0409 said:In how many ways can 4 balls of different colors put in 3 urns which are colored black?
14 ways??
@bs0409 said:In how many ways can 4 balls of different colors put in 3 urns which are colored black?
EDITED ...
A B C D
U U U
4 0 0 -> 4C4 = 1 way
3 1 0 -> 4C3 = 4 ways
2 2 0 -> 4C2/2! = 3 ways
2 1 1 -> 4C2*2C1/2! =6ways
14 ways ?
@bs0409 said:In how many ways can 4 balls of different colors put in 3 urns which are colored black?
14 ways??
@Budokai001 said:@mailtoankit bro, can you check what is wrong with my approach above ?
4 0 0--> 4c4= 1 way
3 1 0-->4c3=4 ways
2 2 0-->4c2/2!=3 ways
2 1 1-->4c2*2c1/2!=6 ways
total 14 ways
PS sahi hai ki nahi ...yeh nahi pata...whats the OA?
@Budokai001 said:@mailtoankit bro, can you check what is wrong with my approach above ?
For case you need to divide by 2!. Rest is correct!!
@mailtoankit said:4 0 0--> 4c4= 1 way3 1 0-->4c3=4 ways2 2 0-->4c2/2!=3 ways2 1 1-->4c2*2c1/2!=6 waystotal 14 waysPS sahi hai ki nahi ...yeh nahi pata...whats the OA?
14 is correct
In how many ways 200 can be represented as
1. sum of consecutive numbers
2. Sum of consecutive even numbers
3.Sum of consecutive odd numbers
In how many ways 200 can be represented as
1. sum of consecutive numbers
2. Sum of consecutive even numbers
3.Sum of consecutive odd numbers
2 1 1-->4c2*2c1/2!=6 ways
Why have you multiplied by 2C1 ??
Lets say uve selected 2(A,B) by 4C2 ... remaining is C,D .. why do you need to select between these two are urns are similar ? ....already out of shape :banghead:
@mailtoankit said:4 0 0--> 4c4= 1 way3 1 0-->4c3=4 ways2 2 0-->4c2/2!=3 ways2 1 1-->4c2*2c1/2!=6 waystotal 14 waysPS sahi hai ki nahi ...yeh nahi pata...whats the OA?
why hv we not considered the case where each urn is empty ??
@pirateiim478 said:For case you need to divide by 2!. Rest is correct!!14 is correct
In how many ways 200 can be represented as
1. sum of consecutive numbers
2. Sum of consecutive even numbers
3.Sum of consecutive odd numbers
Do you mean natural numbers or integers??
@Budokai001 said:2 1 1-->4c2*2c1/2!=6 ways Why have you multiplied by 2C1 ??Lets say uve selected 2(A,B) by 4C2 ... remaining is C,D .. why do you need to select between these two are urns are similar ? ....already out of shape
It means that 2 balls are together (4c2) and other 2 balls are in 2 different boxes. No need to use selection when single ball goes to single box.
@Budokai001 Go through thishttp://totalgadha.com/mod/forum/discuss.php?d=3537 . I was also confused first time!!
@Budokai001 Go through thishttp://totalgadha.com/mod/forum/discuss.php?d=3537 . I was also confused first time!!
@karan20 said:why hv we not considered the case where each urn is empty ??
Question is 4 balls should go into 3 urns. So the case when each urn is not considered as total shld. be 4 finally.
@bs0409 said:Do you mean natural numbers or integers??
Sorry it is for +ve integers only