solve these for me..plz...with detailed solutions

newguy · **Re: solve these for me..plz...with detailed solutions -** 10-09-2004, 02:20 PM

Quote:

Originally Posted by whocarez

Quote:

Originally Posted by cat_cracker1

There is a question
There are four pens each a different colour and four caps each a different colour but,same as that of one of the pens.In how many ways can these caps be put on the puns so that a cap does not go to the pen of same colour.

These prob. Can be done by considering all the cases for c1 going in p2,p3 and p4 and so on.In which the answer is 9.But if the above quantities are increased would it be feasible to go as in the above way.Is there any shortcut method to solve these type of ques.

Yes there is a formula for this type of questions. The type mentioned above is a classic case of dearrangement!!

It should be 4! [1 - 1/1! + 1/2! - 1/3! + 1/4!] which equals 24-24+12-4+1=9

Could you just explain what dearrangement is ?

genie04 · **Re: solve these for me..plz...with detailed solutions -** 10-09-2004, 03:12 PM

this can be done the other way round too..

total ways of capping the pens is:4+3+2+1=10

only one way of capping them correctly.

so total incorrect ones=10-1=9

cheers!

whocarez · **Re: solve these for me..plz...with detailed solutions -** 10-09-2004, 05:46 PM

Quote:

Originally Posted by newguy

Quote:

Originally Posted by whocarez

Quote:

Originally Posted by cat_cracker1

There is a question
There are four pens each a different colour and four caps each a different colour but,same as that of one of the pens.In how many ways can these caps be put on the puns so that a cap does not go to the pen of same colour.

These prob. Can be done by considering all the cases for c1 going in p2,p3 and p4 and so on.In which the answer is 9.But if the above quantities are increased would it be feasible to go as in the above way.Is there any shortcut method to solve these type of ques.

Yes there is a formula for this type of questions. The type mentioned above is a classic case of dearrangement!!

It should be 4! [1 - 1/1! + 1/2! - 1/3! + 1/4!] which equals 24-24+12-4+1=9

Could you just explain what dearrangement is ?

A derangement of n ordered objects, denoted , is a permutation in which none of the objects appear in their "natural" (i.e., ordered) place. For eg. if there are 3 houses numbered 1,2,3 and three owners 1,2,3 of the respective house no. Then classic case of dearrangement in this situation would be that owners can live in any house other than theirs.

dhallsulabh · 10-09-2004, 08:11 PM

excuse me for interrupting...i dont think any of you got the last question right...the caps can go all pens except the pen with same color...so that means there is only 1 combination that will be wrong.....so now total combinations -1 willbe our answer....now for calculatign total no. os cases...we see how many choices we have...so we have 4! choices...which is equal to 24 ...so the answer should be 23 ways.

newguy · 10-09-2004, 09:31 PM

Quote:

Originally Posted by dhallsulabh

excuse me for interrupting...i dont think any of you got the last question right...the caps can go all pens except the pen with same color...so that means there is only 1 combination that will be wrong.....so now total combinations -1 willbe our answer....now for calculatign total no. os cases...we see how many choices we have...so we have 4! choices...which is equal to 24 ...so the answer should be 23 ways.

Buddy ! Read the question carefully "In how many ways can these caps be put on the pens so that a cap does not go to the pen of same colour". The case which you excluded is when all the caps go in the respective color's pen. We can't even allow ONE SINGLE CAP to go in the pen of same color.