@scrabbler said:Nice approach! Mine was much longer...regardsscrabbler
Here it's 3 diff factors but 
Just saw....It should be 40 then?
Just saw....It should be 40 then?@ScareCrow28 said:Here it's 3 diff factors but Just saw....It should be 40 then?
why 40??? 41 is correct only toh... that is what I got... 
.. again, in a very long method... :(
.. again, in a very long method... :(@koyal1990 said:why 40??? 41 is correct only toh... that is what I got... .. again, in a very long method...
Question me 3 diff factors hai..to (1,1,2310) wala nai ayega naa..
@bs0409 said:what is the remainder when 50^56^52 is divided by 11?a-7b-5c-9d-10
5 ?
E(11) = 10
50^10k mod 11 = 1
56^52 mod 10 = 6
50^6 mod 11 = 6^6 mod 11 = 216*216 mod 11 = 7*7 mod 11 = 49 mod 11 = 5
@ScareCrow28 said:Question me 3 diff factors hai..to (1,1,2310) wala nai ayega naa..
oh oh!!! yaaa.. ye nai ayega... 
:banghead:.. then it should be 40... you are absolutely correct!! 
:banghead:.. then it should be 40... you are absolutely correct!! @ScareCrow28 said:Here it's 3 diff factors but Just saw....It should be 40 then?
Oops...yes 😞 Careless, all of us...
regards
scrabbler
regards
scrabbler
@Logrhythm said:@scrabbler ye wala dekho fir zara...In how many ways 2310 be written as product of 3 different factors??
2310 = 2*3*5*7*11
no. of ways = (3^(n - 1) + 1)/2...where n = no. of diff. prime factors
----> 3^(5-1) + 1/2 = 3^4 + 1/2 = 82/2 = 41 ?
no. of ways = (3^(n - 1) + 1)/2...where n = no. of diff. prime factors
----> 3^(5-1) + 1/2 = 3^4 + 1/2 = 82/2 = 41 ?
if a and c are integer and 3how many equation of the form ax^2+12x+c=0 have real roots?
1.36
2.39
3.44
4.48
@bs0409 said:if a and c are integer and 3how many equation of the form ax^2+12x+c=0 have real roots?1.362.393.444.48
44 ??
144 - 4ac >= 0
ac
For, a=3____13 values
a=4____10 values
a=5____8 values
a=6____7 values
a=7____6 values
Total = 44 values ...
@ScareCrow28 said:44 ??144 - ac >= 0acFor, a=3____13 valuesa=4____10 valuesa=5____8 valuesa=6____7 valuesa=7____6 valuesTotal = 44 values ..Shouldn't the condition be 144-4ac>=0
20 children r standing in a line outside a ticket window . 10 of them have one rupee coin each & 10 have two rupee coin each . the entry ticket is priced Rs1. if all arrangements of 2o children are equally likely , find probability that 10th child will be first to wait for change .
The total number of Integral solutions of uvw²x²y² =277200
@jain4444 said:20 children r standing in a line outside a ticket window . 10 of them have one rupee coin each & 10 have two rupee coin each . the entry ticket is priced Rs1. if all arrangements of 2o children are equally likely , find probability that 10th child will be first to wait for change .
0?
regards
scrabbler
regards
scrabbler
@jain4444 said:The total number of Integral solutions of uvw²x²y² =277200
Getting 180....feel like I have missed something....retrying
Edit: I give up. Too much noise in the street outside....some procession...can't concentrate :(
regards
scrabbler
Edit: I give up. Too much noise in the street outside....some procession...can't concentrate :(
regards
scrabbler
@jain4444 said:20 children r standing in a line outside a ticket window . 10 of them have one rupee coin each & 10 have two rupee coin each . the entry ticket is priced Rs1. if all arrangements of 2o children are equally likely , find probability that 10th child will be first to wait for change .
10*10!^2/20! ?
@jain4444 said:20 children r standing in a line outside a ticket window . 10 of them have one rupee coin each & 10 have two rupee coin each . the entry ticket is priced Rs1. if all arrangements of 2o children are equally likely , find probability that 10th child will be first to wait for change .
My take-0..........
@ScareCrow28 said:Is it 41 by any chance?2310 = 2*3*5*7*11No of ways when all factors are diff = (3^5-3)/3!When 2 are same, i.e, 1 = 1 way (1,1,2310)Total = 41 ways ?? ..
@scrabbler said:Ummm....41? aila confusing hai yeh waala!regardsscrabbler
Yes 41 hi hona chaiye...baaki meine abhi office mein solve nahi kara hai...
@koyal1990 said:sir nai samajh aaya.. ek bar phir se samjhao.. aapne manually cases nikale h kya??? if yess then tell me..
ma'am....aapko PM kar dunga....abhi thoda office ka kaam chal raha hai.. :)
@jain4444 said:20 children r standing in a line outside a ticket window . 10 of them have one rupee coin each & 10 have two rupee coin each . the entry ticket is priced Rs1. if all arrangements of 2o children are equally likely , find probability that 10th child will be first to wait for change .
0 ? 
For any child to wait for the first time, there should be as many 2s as 1s, so that all the change gets exhausted till 10th child, but that is not possible since 9 is odd no..
So 0?? Pata ni sir bhot confusion hai...ans btao ..
.. @Logrhythm said:Yes 41 hi hona chaiye...baaki meine abhi office mein solve nahi kara hai... ma'am....aapko PM kar dunga....abhi thoda office ka kaam chal raha hai..
It's "diff factors" in the question, so 40 hona chahiye na