Not very sure what your question means. But I believe that the question means each of the subject is there at least once and one of the subject appears twice.
This subject which appears twice can be selected in 4 ways..
Example AABCD we can arrange this in 5!/2!=60 ways
A question paper is divided into 3 sections-A,B, C having 3,4,5 questions respectively. find no of ways of attempting 6 questions choosing atleast 1 from each section?
@ rahul I have no options I know the answer but dont know how they arrived at it. answer is 12C6-7C6-9C6-8C6=805 ........ I was thinking why dont we do this 3 , 4 , 5 1 , 1 , 1 .....(selecting one from each..3*4*5) 2 , 3 , 4....(remaining) now we can use a+b+c=3 (as 6-3=3) we will get ((3+3-1C2)-1) as "a" cant be 3. so finally the answer will be 3*4*5*9=540 not equal to 805.. Is my approach wrong?
a.15 b.14 c.3 d.None of these ----------------------------------------------- see what i was thinking is using wilsons theorem Rem(17!.18.19.20.21.22)/23= -1 17!= 17!.18.19..22/(18.19.20.21.22) rem= -1 / (-5*-4*-3*-2*-1) rem=-1/(-120)= 1/120 after this i am unable to proceed. Can u help me after this?
4N when divided by D leaves a remainder 9. 3N when divided by D leaves a remainder 13. what is the remainder when N is divided by D?
----------------------------------------------- there is one way by equations. But I had some other thing in my mind 4N leaves 9 as remainder so N should leave 9/4 as remainder Now 3N leaves 13 as remainder so N shd leave 13/4 as remainder so can we arrive at the answer using 13/4 and 9/4?
Given in a triangle ABC, there are four interior points on side AB , four interior points on side BC , & six interior points on side CA . find no. of different lines joining these points.
Dear Rahul If the question had been that there are 4 co linear points , other 4 co linear points and a different set of 6 colinear points in a plane then answer would have been :
14c2-4c2-4c2-6c2+3
But my point is if the points are on sides of triangle then we wont be adding 3 as these three lines are already made.
From a bag containing 34 balls, one ball weighs 4 grams and all the other weigh 5 grams each. Using a simple balance where balls can be kept on either pan, what is the minimum weighs required to identify the defective ball?
please provide a detailed solution to problem of day sep-30-2011 and sep-27-2011
ReplyDeleteYour queries are answered.
ReplyDeleteHope that helps.
Sorry for the delay was out for the weekend.
how many 4 letter words(with repetition) are there such that the letters are there in alphabetical order?
ReplyDeletehow many 4 letter words(without repetition) are there such that the letters are there in alphabetical order?
ReplyDeleteThe solution for the without repetition is easy..
ReplyDeleteFirst take all the words without any contraints, then as we have no repetition.
Then we have 26x25x24x23 words.
Now take for example A, B, C, D they will form 4!=24 words of which only one ABCD will be in order.. Hence 1 word in every 24 suits our requirement..
So, the total no of required words is 26x25x24x23/24=26x25x23= 650x23=14950
For the case where repetition is not allowed, I am not able to figure out a solution which is not something like a brute-force. Give me sometime..
ReplyDeleteno of ways of arranging 4 subjects in 5 periods such that all subjects will be there in the arrangement?
ReplyDeleteNot very sure what your question means. But I believe that the question means each of the subject is there at least once and one of the subject appears twice.
ReplyDeleteThis subject which appears twice can be selected in 4 ways..
Example AABCD we can arrange this in 5!/2!=60 ways
So we have a total of 60x4=240 arrangements.
A question paper is divided into 3 sections-A,B, C having 3,4,5 questions respectively. find no of ways of attempting 6 questions choosing atleast 1 from each section?
ReplyDeleteIf we had options, we could answer this better.. Do you have options?
ReplyDelete@ rahul
ReplyDeleteI have no options
I know the answer but dont know how they arrived at it.
answer is 12C6-7C6-9C6-8C6=805
........
I was thinking why dont we do this
3 , 4 , 5
1 , 1 , 1 .....(selecting one from each..3*4*5)
2 , 3 , 4....(remaining)
now we can use a+b+c=3 (as 6-3=3)
we will get ((3+3-1C2)-1) as "a" cant be 3.
so finally the answer will be
3*4*5*9=540 not equal to 805..
Is my approach wrong?
What they are doing is the right way to do it..
ReplyDeleteI will explain you how .. Let us select any 6 from the 12 questions without any constraints.. then we have 12c6 options..
But we do not want a case where 6 are selected from only two of the sections.. so we need to remove those
which comes only from A and B that is 7c6
only from B and C 9c6
only from A and C 8C6
Please note that it is not possible to select all 6 from 1 section, else we would have needed to check for that and add it back..
Hence the required number is
12c6-7c6-8c6-9c6=805..
Hope that helps..
Do ask if you have any further doubt..
any material on DI and geometry?
ReplyDeleteWhat is the remainder when 17! is divided by 23?
ReplyDeletea.15 b.14 c.3 d.None of these
-----------------------------------------------
see what i was thinking is using wilsons theorem
Rem(17!.18.19.20.21.22)/23= -1
17!= 17!.18.19..22/(18.19.20.21.22)
rem= -1 / (-5*-4*-3*-2*-1)
rem=-1/(-120)= 1/120
after this i am unable to proceed.
Can u help me after this?
4N when divided by D leaves a remainder 9.
ReplyDelete3N when divided by D leaves a remainder 13.
what is the remainder when N is divided by D?
-----------------------------------------------
there is one way by equations.
But I had some other thing in my mind
4N leaves 9 as remainder
so N should leave 9/4 as remainder
Now 3N leaves 13 as remainder
so N shd leave 13/4 as remainder
so can we arrive at the answer using 13/4 and 9/4?
Your method is flawed.. Take for example when N= 5 and D =7
ReplyDelete4N gives a remainder 6 when divided by 7
N gives a remainder 5 when divided by 7.
But 5 is not equal to 6/4.
Remainders do not follow arithmetic laws
22! gives a remainder -1
ReplyDelete17!(18x19x20x21x22) gives a remainder -1
17!(-120) gives a remainder -1
17!x18 gives a remainder 22
now check the odd multiples of 23, reduce them by 1, and find teh one which is a multiple of 18
the answer is 253
which is 23x11
Hence 17! gives a remainder 11.
thanks rahul :)
ReplyDeletehi
ReplyDeleteI was trying to solve Maximum and minimum questions in DI.
They seem really tough.
Any simple material which can help me in that
The following table gives no of students who secures more than 90% in each of 5 subjects from 6 to 10 at a school in 2009.
ReplyDeleteclass/subjects A B C D E
6 12 16 15 22 18
7 15 22 22 21 15
8 07 18 16 23 17
9 10 19 15 22 18
10 15 25 21 29 16
no of students in each class in 2009
6 30
7 35
8 38
9 36
10 40
q1)In class 7 no of students who scored more than 90% in atleast 2 of the 5 subjects is atmost?
a)12 b)15 c)18 d)20
q2)the no of students who scored wmore than 90% in exactly 4 subjects for all the classes put together is atmost?
a)28 b)61 c)96 d)107
q3)no of students in class 6 who scored more than 90% in a maximum of 2 subjects is atmost?
a)19 b)21 c)22 d)26
I don't have much on LR. trying solving past cat questions and if possible the last three years mock questions of CL.
ReplyDeleteGiven in a triangle ABC, there are four interior points on side AB , four interior points on side BC , & six interior points on side CA . find no. of different lines joining these points.
ReplyDeleteDear Rahul If the question had been that there are 4 co linear points , other 4 co linear points and a different set of 6 colinear points in a plane then answer would have been :
14c2-4c2-4c2-6c2+3
But my point is if the points are on sides of triangle then we wont be adding 3 as these three lines are already made.
Please clear me ASAP.
No we would be because AB AC and BC are also to be counted, which get ignored when we subtract the three collinear sets
ReplyDeleteBut here Ab,BC and CA are already drawn and we are just subtracting those lines which will be again drawn by these points overlapping AB,BC and CA.
ReplyDeleteMy point is Triangle is already Drawn . Those 3 sides are already there.
From a bag containing 34 balls, one
ReplyDeleteball weighs 4 grams and all the other
weigh 5 grams each. Using a simple
balance where balls can be kept on
either pan, what is the minimum
weighs required to identify the
defective ball?
Three white squares are selected on a chessboard. Find the number of ways such that no two of them are on same row or column.
ReplyDeleteThanking You,
Arpit Bajpai
Hi...
ReplyDeletePlz guide me regarding cat 2013
Regards
Shiwami Sharma