Three X's, three Y's, and three Z's are placed randomly in a 3 x 3 grid. What is the probability that
no row or column will contain two of the same letter?
@kingsleyx said:Three X's, three Y's, and three Z's are placed randomly in a 3 x 3 grid. What is the probability that no row or column will contain two of the same letter?
6/(9c3*6c3)
=2/(28*20)
=1/280
=2/(28*20)
=1/280
@kingsleyx said:Three X's, three Y's, and three Z's are placed randomly in a 3 x 3 grid. What is the probability that no row or column will contain two of the same letter?
3/(9!/3!*3!*3!)? EDIT
@Torque024 said:Let a, b, c be positive integers, with c > 1.How many solutions are there to the equation a^c ˆ' b^c = 2^100?
0? 
@kingsleyx said:Three X's, three Y's, and three Z's are placed randomly in a 3 x 3 grid. What is the probability that no row or column will contain two of the same letter?
sample space=total number of arrangements=9!/3!*3!/3!
favorable cases =3
==> Probability=1/560
@Brooklyn said:@vijay_chandola : fav case kaise nikala??
take 3 spots from 9 for x - 9c3
take 3 for y - 6c3
9c3*6c3
take 3 for y - 6c3
9c3*6c3
@Brooklyn said:@vijay_chandola : fav case kaise nikala??
Place any letter in first box let be X and then fill accordingly so that no condition violates,
X Y Z
Y Z X
Z X Y
First box can contain any three so 3 ways
Y Z X
Z X Y
First box can contain any three so 3 ways
@Brooklyn said:@vijay_chandola : fav case kaise nikala??
thoda doubtful lg raha h 
Put the x in diagonals. = we can arrange rest 3y and 3z in 2 ways
then, put y in diagonals= 2 cases
put z in diagonals =2 cases.
so there should be 6*2=12 favorable cases (for two diagonals) I guess.
@krum dekhna jara bhai agr kuchh galti h isme to :lookaround:
@Torque024 said:Let a, b, c be positive integers, with c > 1.How many solutions are there to the equation a^c ˆ' b^c = 2^100?
0??
a^n - b^n is always divisible by (a-b)
but then a+b will not be divisible for even n...
and for odd n's number will not come a power of 2...
a^n - b^n is always divisible by (a-b)
but then a+b will not be divisible for even n...
and for odd n's number will not come a power of 2...
@insane.vodka said:0??a^n - b^n is always divisible by (a-b)but then a+b will not be divisible for even n...and for odd n's number will not come a power of 2...
OA is 49
@vijay_chandola said:thoda doubtful lg raha h Put the x in diagonals. = we can arrange rest 3y and 3z in 2 ways then, put y in diagonals= 2 casesput z in diagonals =2 cases.so there should be 6*2=12 favorable cases (for two diagonals) I guess.@krum dekhna jara bhai agr kuchh galti h isme to
sir i think 6 hoga
12 karne par agar rotate kar denge to wahi baat ho jaegi
x-y-z
z-x-y
y-z-x
x-z-y
y-x-z
z-y-x
similarly 2 cases each when y,z are in diagonal
12 karne par agar rotate kar denge to wahi baat ho jaegi
x-y-z
z-x-y
y-z-x
x-z-y
y-x-z
z-y-x
similarly 2 cases each when y,z are in diagonal
@Torque024 said:Let a, b, c be positive integers, with c > 1.How many solutions are there to the equation a^c ˆ' b^c = 2^100?
8?
@krum said:sir i think 6 hoga12 karne par agar rotate kar denge to wahi baat ho jaegi
haan 6 hoga. 😃 y aur z ki positions ko change krne se it would be same 
Because of 2 diagonals, we have to multiply by 2
So, 1/280 then?
@kingsley : What's OA??
@vijay_chandola said:haan 6 hoga. y aur z ki positions ko change krne se it would be same Because of 2 diagonals, we have to multiply by 2So, 1/260 then?@kingsley : What's OA??
Sorry, Dont have the OA. I got 6 fav too. It must be 1/280 then !!
@gautam22 said:sir c=2 leke hi mera 49 aa raha hai jaise jaise c ki value change karenge ye to aur aa jayega....dekhna ek minute main kahn galat kar raha hoon(a-b)(a+b)=2^100......a-b=2^1.......49......a+b=2^99.......51.....isi se 49 values aa rahhi hain
OA 49 hai, mistake ho gai thi.
log 2 is ?
(here base is 5)
a)an integer
b)a rational number
c)a prime number
d)an irrational number