@iLoveTorres said:i mean what's wrong in the logic
Nothing. You took only 1 case out of the 2 (either 3M 1f or 3F 1M...but not both...)
regards
scrabbler
regards
scrabbler
@scrabbler said:Nothing. You took only 1 case out of the 2 (either 3M 1f or 3F 1M...but not both...)regardsscrabbler
how can you choose both the cases particularly in this problem.. wouldnt that make it 4+4=8 persons selection where you would have 2 couples?
@iLoveTorres said:how can you choose both the cases particularly in this problem.. wouldnt that make it 4+4=8 persons selection where you would have 2 couples?
Suppose you had to choose 2 girls or 2 guys from the six couples. How many ways could you do it? Would it be 6C2? or 6C2 + 6C2?
regards
scrabbler
regards
scrabbler
@scrabbler said:Suppose you had to choose 2 girls or 2 guys from the six couples. How many ways could you do it? Would it be 6C2? or 6C2 + 6C2?regardsscrabbler
6C2+6C2
@iLoveTorres said:6C2+6C2
There you have it. That is one problem right there...then you assumed the other 2 people of the same sex, uska bhi kuch locha hai shayad.
regards
scrabbler
regards
scrabbler
@scrabbler said:There you have it. That is one problem right there...then you assumed the other 2 people of the same sex, uska bhi kuch locha hai shayad.regardsscrabbler
so basically i should have considered 3&1 ,2&2 wale cases instead of jus considering 3&1 case?
if thats the case
3M+1F = 6C3*3C1=20*3=60
2M + 1F(couple) + 1F = 6C2*2C1*4C1=15*2*4=120
So you would have total favourable cases as 360
prob = 360/495?
P.S i am not simplifying jus want to kno iif the counting of the cases are right?
if thats the case
3M+1F = 6C3*3C1=20*3=60
2M + 1F(couple) + 1F = 6C2*2C1*4C1=15*2*4=120
So you would have total favourable cases as 360
prob = 360/495?
P.S i am not simplifying jus want to kno iif the counting of the cases are right?
There are 2 brothers among a group of 20 persons. In how many ways can the group be arranged around a circle so that there is exactly one person between the two brothers?
(1) 2 * 17!
(2) 18! * 18
(3) 19! * 18
(4) 2 * 18!
(5) 2 * 17! * 17!
(1) 2 * 17!
(2) 18! * 18
(3) 19! * 18
(4) 2 * 18!
(5) 2 * 17! * 17!
@Dexian said:There are 2 brothers among a group of 20 persons. In how many ways can the group be arranged around a circle so that there is exactly one person between the two brothers?(1) 2 * 17! (2) 18! * 18 (3) 19! * 18 (4) 2 * 18! (5) 2 * 17! * 17!
@Dexian said:There are 2 brothers among a group of 20 persons. In how many ways can the group be arranged around a circle so that there is exactly one person between the two brothers?(1) 2 * 17! (2) 18! * 18 (3) 19! * 18 (4) 2 * 18! (5) 2 * 17! * 17!
18C1 * 2C1 * 17! = 2*18!
@Dexian said:There are 2 brothers among a group of 20 persons. In how many ways can the group be arranged around a circle so that there is exactly one person between the two brothers?(1) 2 * 17! (2) 18! * 18 (3) 19! * 18 (4) 2 * 18! (5) 2 * 17! * 17!
2*18!
@Dexian said:There are 2 brothers among a group of 20 persons. In how many ways can the group be arranged around a circle so that there is exactly one person between the two brothers?(1) 2 * 17! (2) 18! * 18 (3) 19! * 18 (4) 2 * 18! (5) 2 * 17! * 17!
is it 4th option?
fix one person.
make the bros sit on either sides. arrange the rest 18 as u arrange in a circle.
so bros can be arranged in 2! and rest 18 in (n-1)! or 17!
now the person whom u fixed can be in all those 18 places as well
= 2*18*17! = 2*18!
fix one person.
make the bros sit on either sides. arrange the rest 18 as u arrange in a circle.
so bros can be arranged in 2! and rest 18 in (n-1)! or 17!
now the person whom u fixed can be in all those 18 places as well
= 2*18*17! = 2*18!
OA (4) 2 * 18!
@Dexian said:There are 2 brothers among a group of 20 persons. In how many ways can the group be arranged around a circle so that there is exactly one person between the two brothers?(1) 2 * 17! (2) 18! * 18 (3) 19! * 18 (4) 2 * 18! (5) 2 * 17! * 17!
2 * 18! ?
@Dexian said:There are 2 brothers among a group of 20 persons. In how many ways can the group be arranged around a circle so that there is exactly one person between the two brothers?(1) 2 * 17! (2) 18! * 18 (3) 19! * 18 (4) 2 * 18! (5) 2 * 17! * 17!
2*18!
@jain4444 said:How many sets of three distinct factors of the number N = 2^6 — 3^4 — 5^2 can be made such that product of these factors is equal to N and HCF of these factors(in particular set) is equal to 1 ?
Answer is 4?
Guys here is first one for today:
If the equation (y^2 - 2my -4(m^2 + 1))(y^2 - 4y -2m(m^2 + 1)) = 0 has exactly 3 different roots, then how many values can "m" take?
(a) 4 (b) 3 (c) 2 (d) 1
Team BV - Vineet
@bodhi_vriksha said:Guys here is first one for today:If the equation (y^2 - 2my -4(m^2 + 1))(y^2 - 4y -2m(m^2 + 1)) = 0 has exactly 3 different roots, then how many values can "m" take?(a) 4 (b) 3 (c) 2 (d) 1 Team BV - Vineet
Only 1, m=3?
For which the roots can be 0, +10 and -10... Just tried trial and error
Another one:
For all real k, f(k) satisfies 2f(k) + f(1-k) = 2*k^2 + 1. Then which among the following is
true for all k?
(a) f(k) = -1 (c) f(k) >= 1 (d) f(k)
Team BV - Vineet
@bodhi_vriksha said:Another one:For all real k, f(k) satisfies 2f(k) + f(1-k) = 2*k^2 + 1. Then which among the following istrue for all k?(a) f(k) = -1 (c) f(k) >= 1 (d) f(k) Team BV - Vineet
option b