Seven different objects must be divided among 3 ppl. In how many ways can this be done if atleast one of them gets exactly 1 object??
@jain4444 said:How many pairs of integers (not necessarily positive) are there such that a^2 + 6b^2 is squares (a
6a^2+b^2=x^2
x^2-b^2 has to be a multiple of 6
since only odd nos and multiples of 4 can be expressed as difference of squares
x^2- b^2 has to be a multiple of 4
6*2 6*4 ... are possible
but since 2 6 .. are not squares we have to remove them
6*4
6*16
..
6*256
hence 4??
@ani6 said:Seven different objects must be divided among 3 ppl. In how many ways can this be done if atleast one of them gets exactly 1 object??
1176 ?
1 2 3
5 1 1----> 7c5*2c1*3!/2 = 126
4 2 1----> 7c4*3c2*3! = 630
3 3 1----> 7c3*4c3*3!/2 = 420
total = 630 + 126 + 420 = 1176
@ani6 said:Seven different objects must be divided among 3 ppl. In how many ways can this be done if atleast one of them gets exactly 1 object??
115 =7c1*6c1/2 *3!=21*6=126
124=7c1*6c2 *3!=7*15*6=630
133=7c1*6c3/2 *3!=420
126+630+840+630=1176
@Subhashdec2 said:115 =7c1*6c1/2 *3!=21*6=126124=7c1*6c2 *3!=7*15*6=630133=7c1*6c3/2 *3!=840223=7c2*5c2/2 *3!=630126+630+840+630=2226
223 waala thode na lenge...exactly one poocha hai ??
Q . thr r t taps 1 , 2 ...t each of which can fill a cistern . the rate of filling of nth tap is such tht it is equal to twice tht of all the taps frm 1 to (n-1) put together . if the 18th tap can fill the empty cistern in 2 mins , then find the time in which the 15th tap alone can fill the empty cistern .
need soln plz .
@jain4444 said:How many pairs of integers (not necessarily positive) are there such that a^2 + 6b^2 is squares (a
a^2 + 6b^2 = k^2
b^2 = k^2-a^2/6
(k^2-a^2) is either of the form 4k or odd...but it cannot be odd since it is also of the form 6k..
so it must be of the form (2*3)*(2^2*x) where x is even...
x=0,2,4,6,8,10,..,30
so 15 solutions??
@jain4444 said:How many pairs of integers (not necessarily positive) are there such that a^2 + 6b^2 is squares (ab has to be an even number to satisfy thisa,b =(0,3) , (1,2) , (-1,2) , (2,4) , (-2,4) ........449,998so 449*2+1=999 pairs of integers999
@Subhashdec2 said:6a^2+b^2=x^2x^2-b^2 has to be a multiple of 6since only odd nos and multiples of 4 can be expressed as difference of squaresx^2- b^2 has to be a multiple of 46*2 6*4 ... are possiblebut since 2 6 .. are not squares we have to remove them6*46*16..6*256hence 4??
a = + - 1 , b = 2
a = + - 2 , b = 4
a = + - 4 , b = 8
a = + - 8 , b = 16
a = + - 16 , b = 32
a = + - 32 , b = 64
a = + - 64 , b = 128
a = + - 128 , b = 256
a = + - 256 , b = 512
a = + - 4 , b = 8
a = + - 8 , b = 16
a = + - 16 , b = 32
a = + - 32 , b = 64
a = + - 64 , b = 128
a = + - 128 , b = 256
a = + - 256 , b = 512
18 solutions in total
If we have to make 7 boys sit alternately with 7 girls around a round table which is numbered then the no of ways in which this can be done is
@ani6 said:If we have to make 7 boys sit alternately with 7 girls around a round table which is numbered then the no of ways in which this can be done is
7!*7!*2? 
@ani6 said:If we have to make 7 boys sit alternately with 7 girls around a round table which is numbered then the no of ways in which this can be done is
7! * 7! ??
@Abir1103 said:Q . thr r t taps 1 , 2 ...t each of which can fill a cistern . the rate of filling of nth tap is such tht it is equal to twice tht of all the taps frm 1 to (n-1) put together . if the 18th tap can fill the empty cistern in 2 mins , then find the time in which the 15th tap alone can fill the empty cistern .need soln plz .
Is the ans 54 minutes?
@ani6 said:If we have to make 7 boys sit alternately with 7 girls around a round table which is numbered then the no of ways in which this can be done is
7!*7!*2 ?(edited)...PnC...
@ani6 said:If we have to make 7 boys sit alternately with 7 girls around a round table which is numbered then the no of ways in which this can be done is
EDIT:
7!*7!*2 ?
Suppose The boys are seated at 1, 3, 5....13 positions now girls have 7! ways
Now if the boys are at 2, 4, 6...14 positions the girls have 7! ways
Total = 7!*7!*2 ..
@jain4444 said:a = + - 1 , b = 2a = + - 2 , b = 4
a = + - 4 , b = 8
a = + - 8 , b = 16
a = + - 16 , b = 32
a = + - 32 , b = 64
a = + - 64 , b = 128
a = + - 128 , b = 256
a = + - 256 , b = 51218 solutions in total
bhai zra btana
(1,2) similarly -1,2
2,4 -2,4
3,6 -3,6
4,8 -4,8
5,10 -5,10
. .
. .
. .
499,998 -499,998
ye sab satisfy karte h yar for x as square
a^2+6b^2 =
@jain4444 said:How many pairs of integers (not necessarily positive) are there such that a^2 + 6b^2 is squares (abhai zra btana0,3(1,2) similarly -1,22,4 -2,43,6 -3,64,8 -4,85,10 -5,10. .. .. .499,998 -499,998ye sab satisfy karte h yar for x as squareto answer 999 hona chahiye 18 kaise ho rha h ??????
@abhishek.2011 said:bhai zra btana(1,2) similarly -1,22,4 -2,43,6 -3,64,8 -4,85,10 -5,10. .. . . .499,998 -499,998ye sab satisfy karte h yar for x as square a^2+6b^2 =
correct hai bhai 
approach batao
P.S - this is not a original question...ye self made hai