@rubber said:How we do these type of problem?
selecting 25 objects out of 15 of 1 kind,20 of 2nd kind and 25 of 3rd kind???
I missed it tooo :(
@[450012:rajatt1557]
hey did u find out how to do this:
There are five balls of different colours and 5 boxes of the same colours as those of the balls.No. of ways in which the balls one in each box can be placed such that the ball does nt goes to the box of its own colour is
@[583666:iim18]
yep. principle of derangement is the answer
@[450012:rajatt1557]:thanks:)
@[450012:rajatt1557] --- Concept of dearrangement --- answer is 44. pls. confirm if its correct..
@[169132:naga25french] ---
Possible cases ...
4!/3!=4
4!/2!=12
4!/2!=12
4!/=6
4!/4!=1
total=35..
Can pls explain how did u get 6??
@[361226:ayushbhalotia]
its correct bro
How we do these type of problem?
selecting 25 objects out of 15 of 1 kind,20 of 2nd kind and 25 of 3rd kind???
@ayushbhalotia said:How we do these type of problem?
selecting 25 objects out of 15 of 1 kind,20 of 2nd kind and 25 of 3rd kind???
i am not sure about this..but since the no of objects =25 and if we select say x,y,z objects of 1st,2nd and 3rd kind then x+y+z=25.the no of postive integral solns=27c2.we have to remove out of this the cases where z>25(not possible according to prb),y>20 and x>15 which are 0,6c2 and 11c2 respectively.we also see that both x and y cannot simultaneously be greater than 15 and resp.so the ans should be 27c2-6c2-11c2=281.but i am not sure of this..what is the answer?
@[607369:nfsfreak] --- Answer is 281 only...
but i did not understand y 27C2...from where 27 came from and y 2??? y 6C2 Y 11C2...?? why 2?? we are not choosing 2 balls out of 1 kind...
'concept derangement' : can anyone tell more about it. If already said, can anyone quote that reply?
@ayushbhalotia said: @nfsfreak --- Answer is 281 only...but i did not understand y 27C2...from where 27 came from and y 2??? y 6C2 Y 11C2...?? why 2?? we are not choosing 2 balls out of 1 kind...
see this sum..here is the link..we are using the idea the no of positve integral solutions
of x1+x2+.....xr=N is N+r-1Cr-1.http://www.pagalguy.com/forums/quantitative-ability-and-di/quant-arun-sharma-t-23813/p-794829/r-2935084. see the solution given to this prb "In an examination the maximum marks for each of the 3 papers are 50 each.Maximum marks for the fourth paper are 100. Find the number of ways in which the candidate can score 60% marks in the aggregate?" .u willl get the idea.
From 20 consecutive natural nos 1-20, 4 nos are been selected. How many ways these 4 nos be selected such that the difference of any 2 nos. selected is more than 3??
Pls solve this sum..
@[415535:mailmesant]--- no ya...answer is 14C4
Q1) 3 nos are selected from the 1st 20 Natural Nos. find the prob that the 3 nos selected are in GP??
How many four digit od numbers can be formed such that every 3 in the no. is followed by 6?
a)108
b)2592
c)2696
d)2700
@ayushbhalotia said: Q1) 3 nos are selected from the 1st 20 Natural Nos. find the prob that the 3 nos selected are in GP??
is it 9/1140????
@[450012:rajatt1557] - in the questn 3 should be followed by 6 means 36 or 63...the no of cases change accordingly...
@[103935:chandrakant.k] --- now u have taken cases as 1,2,4 etc etc...there are 9 cases in which the r > 1...and there are again 9 cases in which r