@hatemonger said:solve this
option 3?
log(p^1/2 *q^2) = 1-1/4
p^2*q^8=8
log base 2 h.
log(p^1/2 *q^2) = 1-1/4
p^2*q^8=8
log base 2 h.
@Asfakul said:Why are you discarding the cases when power of 2 number is the same ? It's not said that the numbers needs to be different ? please explain
see, we just need to remove the redundant cases...
the 78 cases contains ordered triplets hence making duplicate cases...
in those 78 cases -> (1,1,9) is counted thrice as (1,1,9);(9,1,1) and (1,9,1) which should actually be counted only once....isliye eliminate karne hai...
hope it is clear... :)
@The_Loser said:explain.
11! * 7! = a * b
2^12 . 3^6. 5^3 . 7^2. 11 = a*b
number of prime factors = 5
number of pairs of a,b possible = 2^(5-1) = 16
@Logrhythm said:see, we just need to remove the redundant cases...the 78 cases contains ordered triplets hence making duplicate cases...in those 78 cases -> (1,1,9) is counted thrice as (1,1,9);(9,1,1) and (1,9,1) which should actually be counted only once....isliye eliminate karne hai... hope it is clear...
thanks very clear .. 
@jain4444 said:hmm did mistake chalo ye try karo A jar is filled with red, white, and blue tokens that are same but their color. The chance of randomly selecting a red token, replacing it, and then randomly selecting a white token is the same as the chance of randomly selecting a blue token. If the number of tokens of every color is a multiple of 3, what is the smallest possible total number of tokens in the jar?via - kamal/vineet sir
9 ?
@The_Loser said:LCM and HCF of two numbers are 11! and 7! respectively. How many pairs of such numbers are possible?
LCM*HCF = N1 * N2
N1 * N2 = 2^12 *3^6 * 5^3 * 7^2 * 11
Pairs = 2^(5-1) = 16
@The_Loser said:LCM and HCF of two numbers are 11! and 7! respectively. How many pairs of such numbers are possible?
a = 7!*n
b = 7!*m, where n and m are coprime
b = 7!*m, where n and m are coprime
a*b = LCM*GCD
n*m = 8*9*10*11
8*9*10*11 has exactly 4 prime factors, so number of ways in which 8*9*10*11 can be written as product of two co-prime numbers is (2^4)/2 = 8
Hence, there will be 8 such pairs
@Asfakul said:In how many ways can 7^11 be expressed as a product of 3 numbers?Please explain in detail .
x+ y + z=11
13c2
but here as we need pairs so we need unordered pairs
2 same = 6 cases
13c2 - 6*3/3! = 60/6 =10
10+ 6 = 16
@hatemonger said:solve this
2) option 3?
p^2*q^8=8?
log(2)p^1/2+log(2)q^2=1+(-1/4)log(2)2
log(2)[p^1/2*q^2]=1-1/4=3/4
p^1/2*q^2=2^3/4
raise to the 4th power
p^2*q^8=2^3=8
p^2*q^8=8?
log(2)p^1/2+log(2)q^2=1+(-1/4)log(2)2
log(2)[p^1/2*q^2]=1-1/4=3/4
p^1/2*q^2=2^3/4
raise to the 4th power
p^2*q^8=2^3=8
@chillfactor said:a = 7!*nb = 7!*m, where n and m are coprimea*b = LCM*GCDn*m = 8*9*10*118*9*10*11 has exactly 4 prime factors, so number of ways in which 8*9*10*11 can be written as product of two co-prime numbers is (2^4)/2 = 8Hence, there will be 8 such pairs
your answer is correct, but I am not getting the right way.
why din you pic up entire 7! & 11! as whole. why just only 8,9,10,11. ?
why din you pic up entire 7! & 11! as whole. why just only 8,9,10,11. ?
@The_Loser said:LCM and HCF of two numbers are 11! and 7! respectively. How many pairs of such numbers are possible?
ans :8
ans :8
@abhishek.2011 said:x+ y + z=1113c2but here as we need pairs so we need unordered pairs2 same = 6 cases13c2 - 6*3/3! = 60/6 =1010+ 6 = 16
Bhai why 60/6 ?
test test.........