@scrabbler Can you plz elaborate how you got 65 for K?
@bodhi_vriksha said:We need to choose two of 32 whites AND choose 1 of 32 blacks.32C2 * 32C1Now solve this:In how many ways three white squares can be selected on a chessboard such that no two squares are in same row or column?
32C1*25C1*18C1/(3!)
@bodhi_vriksha said:This is not correct. Please try again.Think in terms of positioning of the subsequent left over squares.
Ah sorry it will be
32*(16*18 + 9*16)/6 = 2304 ways
32*(16*18 + 9*16)/6 = 2304 ways
@chillfactor said:Ah sorry it will be32*(16*18 + 9*16)/6 = 2304 ways
Still incorrect, but now you are thinking in the right direction. Come on nail it in your next try...
@bodhi_vriksha said:Still incorrect, but now you are thinking in the right direction. Come on nail it in your next try...
counting mistake
it will be 32(16*18 + 9*20)/6 = 2496
@iLoveTorres said:for the three different cases i am getting WWB 32c1*25c1*13c1/2BWW 32c1*24c1*6c1/2WBW 32c1*24c1*9c1/2Whats the error here?
How are you getting 13, 6, or 9 cases for last square. Something is missing here.
@grkkrg said:6912?Choose a black square : 32C1Choose a white square : (32 - 8)C1 = 24C1Choose another white square : (23- (8 - 1)) = 18C1(Removing 1 because there will be one common white with a row or column of both black and white squares selected.)Total = 32 * 24 * 18 / 2 = 6912
Bingo!! :thumbsup:
@chillfactor said:How are you getting 13, 6, or 9 cases for last square. Something is missing here.Bingo!!
sorry they should be 16,9,12..
Now what is my mistake?
Now what is my mistake?
There are six friends writing the TAC test for admission to nine MIIs. In how many ways could their final results be achieved ??
@jain4444 said:
let f be a function defined on the set of all integers, and assume that it satisfies the following properties:A. f(0) (not equal) 0B. f(1)=3C. f(x)f(y)=f(x+y) + f(x-y) for all integers x and y.determine f(7).
843 ?
f(1)f(0) = f(1) + f(1)
f(0) = 2
f(1)f(1) = f(2) + f(0)
f(2) = 7
f(2)f(1) = f(3) + f(1)
f(3) = 18
f(3)f(1) = f(4) + f(1)
f(4) = 47
f(3)f(4) = f(7) + f(1)
f(7) = 843
@iLoveTorres said:sorry they should be 16,9,12..Now what is my mistake?
Check this post:-
http://pagalguy.com/forums/quantitative-ability-and-di/official-quant-thread-cat-2012-part-4-t-82184/p-3509369/r-3536166
http://pagalguy.com/forums/quantitative-ability-and-di/official-quant-thread-cat-2012-part-4-t-82184/p-3509369/r-3536166
@chillfactor said:counting mistakeit will be 32(16*18 + 9*20)/6 = 2496
and that's correct :)
Basically we can observe the following tree wrt positions of White squares in the alternate rows/columns:
All whites can be divided into two groups of 16 let's say A and B.
So, if in first go the square from A is selected, there will be 16-7=9 squares left in group A, while there will still be 16 in B. In the second go, we can chose the square from any on A or B. If we select from A, then 9-5=4 squares will be left in A and 16 in B, while if we select from B, then 16-7=9 will be left in B and 9 will also be left in A from the previous step.
This can be seen as 32 AND {(9 AND (4 or16)) OR (16 AND (9 or 9))}, which translates in terms of calculation as (1/3!) *32*(9*(4+16) + 16*(9+9)) =2496
@chillfactor said:Oops sorry my bad! I meant to ask thatIn how many ways one can choose two white squares and one black square in chessboard such that no two lies in same row or column ?
Use the same sort of logic that i explained in the previous post to get the following three cases:
WWB -> (1/2!) 32*(9*16+16*18)
BWW -> (1/2!) 32*(24*18)
WBW -> (1/2!) 32*(24*18)
Seems like Junta is having Friday fun in office, or may be the topic is boring. So, let's change the topic...i know you guys love remainders....
Find the last three digits of 3^1994.
@bodhi_vriksha said:Seems like Junta is having Friday fun in office, or may be the topic is boring. So, let's change the topic...i know you guys love remainders....Find the last three digits of 3^1994.
369
sry didnt see.. read two places. :(
3^1994 / 1000
9. (81)^498
9.41... 369?
sry didnt see.. read two places. :(
3^1994 / 1000
9. (81)^498
9.41... 369?
Another quick one:
What is the remainder when 2^2001 is divided by 2001?
@bodhi_vriksha said:Seems like Junta is having Friday fun in office, or may be the topic is boring. So, let's change the topic...i know you guys love remainders....Find the last three digits of 3^1994.
569 ??
@bodhi_vriksha said:Seems like Junta is having Friday fun in office, or may be the topic is boring. So, let's change the topic...i know you guys love remainders....Find the last three digits of 3^1994.
369?