@jain4444 said:The no of ways of dividing 15 men and women into 15 couples each consisting of man and woman is
bhai total no. of people are 15 ?
@jain4444 said:Divya and Raveena can do a work alone exactly in 20 and 25 days respectively. However, when they work together, they do 25% more work than is expected. If they work for a few days alone and for few days together (both being integers only), then the work could not have been completed in exactly (1) 10 days (2) 14 days (3) 16 days (4) 17 days (5) either none or at least 2 of these
14
@jain4444 said:The no of ways of dividing 15 men and women into 15 couples each consisting of man and woman is
15! in anyway
@viewpt said:15! in anyway
No,It is like this
1m and 1w can be chosen among 15m and 15w is 15c1*15c1=225
again 1m and 1w can be chosen among 14m and 14wis 14c1*14c1=196
.again 1m and 1w can be chosen among 13m and 13wis 13c1*13c1=169
.
.
similarly 1m and 1w can be chosen among 1m and 1w is 1c1*1c1=1
sum of squares till 15 which is ÎŁn^2 (where n=15) = 1240
1m and 1w can be chosen among 15m and 15w is 15c1*15c1=225
again 1m and 1w can be chosen among 14m and 14wis 14c1*14c1=196
.again 1m and 1w can be chosen among 13m and 13wis 13c1*13c1=169
.
.
similarly 1m and 1w can be chosen among 1m and 1w is 1c1*1c1=1
sum of squares till 15 which is ÎŁn^2 (where n=15) = 1240
@IIMAIM said:No,It is like this1m and 1w can be chosen among 15m and 15w is 15c1*15c1=225again 1m and 1w can be chosen among 14m and 14wis 14c1*14c1=196.again 1m and 1w can be chosen among 13m and 13wis 13c1*13c1=169..similarly 1m and 1w can be chosen among 1m and 1w is 1c1*1c1=1 sum of squares till 15 which is ÎŁn^2 (where n=15) = 1240
can you suggest why following is not right way:
1 man can go with one among 15 so 15 choices
2nd cand go with one among 14 so 14 choices..
sililarly
15*14*13*........*1= 15! ??
1 man can go with one among 15 so 15 choices
2nd cand go with one among 14 so 14 choices..
sililarly
15*14*13*........*1= 15! ??
@IIMAIM said:take the same case with 2m and 2wnow according to you 2! wayslets take A,B as men and C,D as womennow (A,C),(A,D),(B,C),(B,D) 4 ways which is not true with your way
Hmm point is thr:
as per your way:
its 2C1*2C1=4
now 1C1* 1C1=1
so 5?? kuch galbad hai??
as per your way:
its 2C1*2C1=4
now 1C1* 1C1=1
so 5?? kuch galbad hai??
@IIMAIM said:take the same case with 2m and 2wnow according to you 2! wayslets take A,B as men and C,D as womennow (A,C),(A,D),(B,C),(B,D) 4 ways which is not true with your way
I think I didn't understand the question.
Men -> A, B
Women -> C,D
First way : A paired with C, B paired with D
Second way: A paired with D, B paired with C..
That's it, how come 4 ways ? 😞
It is asking for number of ways in which they can be divided into 15 pairs, and not the number of distinct pairs which are formed
Number of distinct pairs = 4 in above case
@IIMAIM said:Check @YouMadFellow's comment , I think you might have understood by now.
Arey, I have a doubt 
I am supporting 15! 😞
😞 Make the men stand in a straight line, and keep shuffling the women in front of them, for each arrangement we get a way to divide the 15 men and women into 15 couples.
@YouMadFellow said:Arey, I have a doubt I am supporting 15!Make the men stand in a straight line, and keep shuffling the women in front of them, for each arrangement we get a way to diving the 15 men and women into 15 couples.
1m and 1w can be chosen among 15m and 15w is 15c1*15c1=225
again 1m and 1w can be chosen among 14m and 14wis 14c1*14c1=196
.again 1m and 1w can be chosen among 13m and 13wis 13c1*13c1=169
.
.
similarly 1m and 1w can be chosen among 1m and 1w is 1c1*1c1=1
sum of squares till 15 which is ÎŁn^2 (where n=15) = 1240
did u find anything wrong in this?
again 1m and 1w can be chosen among 14m and 14wis 14c1*14c1=196
.again 1m and 1w can be chosen among 13m and 13wis 13c1*13c1=169
.
.
similarly 1m and 1w can be chosen among 1m and 1w is 1c1*1c1=1
sum of squares till 15 which is ÎŁn^2 (where n=15) = 1240
did u find anything wrong in this?
@YouMadFellow said:Arey, I have a doubt I am supporting 15!Make the men stand in a straight line, and keep shuffling the women in front of them, for each arrangement we get a way to diving the 15 men and women into 15 couples.
bhai am obliged ..IIM jaane wala ek banda toh support kar rha hai...kuch kuch.. @jain4444 bhai ab aa kar aap bhi thodi confusion dur kar do if possible.
@IIMAIM Before deciding wrong and right, I need to understand the question first. What is it asking ? Unique pairs of men and women OR ways to divide them into 15 couples ?
Also, your method should work for 2m, 2w too.
So, in that case, you are getting 2^2 + 1^2 = 5 ? What is this 5 ? 😞 Show me
A,B | C,D -> Which 5 pairs or ways ?
the digits 1, 2 , 3, ....,9
are written at random to form a Nine-digit number..... The probability that it is divisible by "11" is
are written at random to form a Nine-digit number..... The probability that it is divisible by "11" is
@hari_bang said:solve this series...4,20,24,6,2,8...remb only 2 optn...other i forgt..1618...
Next number in the series should be 12....
as it follows the series as:
4,4*5(20),20+4(24),24/4(6),6-4(2),2*4(8),8+4(=12) [alternate +4 and -4 ]
Correct me if I am wrong...
@viewpt said:bhai am obliged ..IIM jaane wala ek banda toh support kar rha hai...kuch kuch.. @jain4444 bhai ab aa kar aap bhi thodi confusion dur kar do if possible.
Arey, I am weak in Permutation, Probability 
..
.. I think, question needs to be understood first, usi me main problem ho rahi hai , aur main abhi IIM se andar nahi gaya hu :splat:
@YouMadFellow said:@IIMAIM Before deciding wrong and right, I need to understand the question first. What is it asking ? Unique pairs of men and women OR ways to divide them into 15 couples ?Also, your method should work for 2m, 2w too. So, in that case, you are getting 2^2 + 1^2 = 5 ? What is this 5 ? Show me A,B | C,D -> Which 5 pairs or ways ?
yeah it work with 2m and 2w too
The question is The no of ways of dividing 15 men and women into 15 couples each consisting of man and woman
Since according to my case A,B are men and C,D are women
Now 2c1*2c1=4 which is the number of ordered pairs i have mentioned
The question is The no of ways of dividing 15 men and women into 15 couples each consisting of man and woman
Since according to my case A,B are men and C,D are women
Now 2c1*2c1=4 which is the number of ordered pairs i have mentioned
@IIMAIM said:yeah it work with 2m and 2w too The question is The no of ways of dividing 15 men and women into 15 couples each consisting of man and womanSince according to my case A,B are men and C,D are womenNow 2c1*2c1=4 which is the number of ordered pairs i have mentioned
for 15 couples you're doing 1^2+2^2+.....+15^2
so for 2 couples you should also do 1^2+2^2 = 5 and not 4.
and 5 is wrong.
so the answer should be 15!
We have look into which man gets which woman.
1st man get women in 15 ways
2nd man gets in 14 ways
..and so on. so we get 15x14x....1= 15!
PS: If we are asked not to arrange the order then there is only ONE way we can group 15men with 15 women since each gets one. But the questions asks the order.
so for 2 couples you should also do 1^2+2^2 = 5 and not 4.
and 5 is wrong.
so the answer should be 15!
We have look into which man gets which woman.
1st man get women in 15 ways
2nd man gets in 14 ways
..and so on. so we get 15x14x....1= 15!
PS: If we are asked not to arrange the order then there is only ONE way we can group 15men with 15 women since each gets one. But the questions asks the order.
@IIMAIM said:bhai in my case it is working when 2m and 2w are there 2c1*2c1=4and after selecting one ordered pair you are left with 1m and 1w so you have to consider that too so total 4+1=5
correct thats what its not comming as 4
@viewpt said:correct thats what its not comming as 4
yeah lets take the case of A,B as men and C,D as women
Now to form couples 1st select 1 man from 2 men and 1 women from 2 women
which is 2c1*2c1=4
and not 5 as the 1 man and 1 women are left (that is treated as different case)
P.S sorry for messing up
Now to form couples 1st select 1 man from 2 men and 1 women from 2 women
which is 2c1*2c1=4
and not 5 as the 1 man and 1 women are left (that is treated as different case)
P.S sorry for messing up