[shashank3012]

@bangaram
Hi,i solved it by making a table of boys vs girls.
From the question,we can see that B2 and B3 have no problems with any girl and so,they are kept for the end.B5 and B6 are dependent on the results of B1 and B4.
Now,B1 can only be with G2,G3 and G6.Consider these cases one by one.
For B1G2,there can be 3 possibilities for B4.They are B4G3,B4G5 and B4G6.Out of these,the G3 and G6 are similar.So,you can multiply by 2.For case 1,you can match B5 and B6 with 3 girls.So,6 possibilities.Also,the last 2 girls can be arranged with B2 and B3 in 2 ways.So,total of 12 ways.Similarly,for the B4G5 case,there are 4*3*2=24 possibilities.And for B4G6 there are 12 possibilities.
So,48 possibilities.
Similarly,for B1G3 there are 36 possibilities and for B1G6 there are 36 possibilities,
So,total is 48+36+36=120 ways of arrangement.
Making a table will help you in understanding this sum.
Hope that helps.

[shashank3012]

For question 5,
The sets can be 1 and 9 other nos. So,9 sets.
With 2,four other nos.so,4 sets
With 3,five other nos.so,5 sets
With 4,three other nos.so,3 sets
With 5,four other nos.so,4 sets
With 6,one no.so,1 set
With 7,three nos.so,3 sets
With 8,one no.so,1 set
With 9,one no. So,1 set
So,total of 31 sets.Also,the order can be changed so,total number of sets are 62.Is this right?

[masoom]

Originally Posted by shashank3012

For question 5,
The sets can be 1 and 9 other nos. So,9 sets.
With 2,four other nos.so,4 sets
With 3,five other nos.so,5 sets
With 4,three other nos.so,3 sets
With 5,four other nos.so,4 sets
With 6,one no.so,1 set
With 7,three nos.so,3 sets
With 8,one no.so,1 set
With 9,one no. So,1 set
So,total of 31 sets.Also,the order can be changed so,total number of sets are 62.Is this right?

i think there is some doubt..

the set here can contain only 2 elements..

[shashank3012]

Originally Posted by masoom

i think there is some doubt..

the set here can contain only 2 elements..

Yeah so the answer should be 31.Is this correct?

[masoom]

Originally Posted by shashank3012

Yeah so the answer should be 31.Is this correct?

dude you just explain the method.. so there will be some discussion..

ans to nikal hi jayega...

[shashank3012]

Now,1 is coprime to each and every integer.So 9 sets can be formed with (1,x),where x can be 2-10.
Similarly,(2,3)(2,5)(2,7)(2,9)
(3,4)(3,5)(3,7)[3,8](3,10)(4,5)
(4,7)(4,9)(5,6)(5,7)[5,8](5,9)
(6,7)[7,8](7,9)(7,10)(8,9)and
(9,10).
So,total of 31 possible sets.

[masoom]

Originally Posted by shashank3012

Now,1 is coprime to each and every integer.So 9 sets can be formed with (1,x),where x can be 2-10.
Similarly,(2,3)(2,5)(2,7)(2,9)
(3,4)(3,5)(3,7)[3,8](3,10)(4,5)
(4,7)(4,9)(5,6)(5,7)[5,8](5,9)
(6,7)[7,8](7,9)(7,10)(8,9)and
(9,10).
So,total of 31 possible sets.

yups the ans is right... i just want to enquire that does (1,x) where x is any number is a co-prime number.. as 1 is neither prime nor composite..

[shashank3012]

I had confirmed the fact before posting the answer ;-).Just got confused by the (a,b) part and so,multiplied by 2 was involved too much in the earlier boys/girls problem.
Source-
Coprime - Wikipedia, the free encyclopedia

[bangaram]

Q) In a triangle ABC, F, E, D are the feet of perpendiculars drawn from a point P inside the triangle on sides AB, BC and CA respectively. If DA = 6 units, AF = 3 units, FB = 4 units, BE = 5 units and EC = 3 units, then find the length of CD.

[shashank3012]

Is the answer sqrt(27) or
3sqrt(3)?If correct i will post the answer.