nforninad SaysFind the no of ways in which 10 different balls can be placed in 4 different boxes such that at max 2 boxes are empty..
Is it
4c2*((2^10)-2)) + 4c3*((3^10)-3)) + ((4^10)-4) ???
Is it
4c2*((2^10)-2)) + 4c3*((3^10)-3)) + ((4^10)-4) ???
nforninad SaysFind the no of ways in which 10 different balls can be placed in 4 different boxes such that at max 2 boxes are empty..
Number of ways in which no box is empty is given by:-
4^10 - 4*3^10 + 6*2^10 - 4 = 818520
Number of ways in which exactly one box is empty is given by:-
4(3^10 - 3*2^10 + 3) = 223920
Number of ways in which exactly two boxes are empty is given by:-
6(2^10 - 2) = 6132
=> Total = 1048572
Alternatively:-
Total cases = 4^10
Number of cases when three are empty = 4
=> 4^10 - 4 = 1048572 cases
Number of ways in which no box is empty is given by:-
4^10 - 4*3^10 + 6*2^10 - 4 = 818520
Number of ways in which exactly one box is empty is given by:-
4(3^10 - 3*2^10 + 3) = 223920
Number of ways in which exactly two boxes are empty is given by:-
6(2^10 - 2) = 6132
=> Total = 1048572
Understood where I went wrong. But, why this plus sign? Didn't get that part.
Understood where I went wrong. But, why this plus sign? Didn't get that part.
Well..actually I unnecessarily complicated it. Just read the alternative solution, its much simpler.
When none pf them gets 0, then the cases will be
4^10 - 4*3^10 + 6*2^10 - 4
Total should be 4^10, but here we counted those cases also when some of the boxes are empty.
So, 4*3^10 are the cases when at least one box is empty, but here also we have over-counted those cases when two or more boxes are empty.
So, C(4, 2)*2^10 is the number of cases when at least two boxes are empty, but here also we over-counted those cases when three boxes are empty.
So, finally there are 4 cases when three boxes are empty.
So, result should be 4^10 - (4*3^10 - (6*2^10 - 4)) = 4^10 - 4*3^10 + 6*2^10 - 4
Note:- Its similar to Principle of Inclusion-Exclusion used in deriving formula for the case of derangements
The AMS MOCK CAT test CATALYST 19 consists of 4 sections. Each section has a maximum of 45 marks. Find the number of ways in which a student can qualify in the AMS MOCK CAT if the qualifying marks is 90.
1) 36546
2) 6296
3) 64906
4) n.o.t
A+B+C+D = 90 0 A,B,C,D 45
Considering lower limit, 93C3
Considering Upper limit 3*47C2
93C3-3*47C2
In a certain examination of 6 papers consisting of max.100 marks each, the number of ways a student can obtain 40% of the overall marks is
C (245,5) - 6*C (144,5) + 15*C (43,5)
I understood how the first two terms came, I can't make out how the "+" term came! help
thanks
In a certain examination of 6 papers consisting of max.100 marks each, the number of ways a student can obtain 40% of the overall marks is
C (245,5) - 6*C (144,5) + 15*C (43,5)
I understood how the first two terms came, I can't make out how the "+" term came! help
thanks
its a use of multinomial theorem:-
We have to find the coefficient of x^240 in (1 + x + x^2 + ... + x^100)^6
= Coeff of x^240 in (1 - x^101)^6 * (1 - x)^(-6)
= Coeff of x^240 in (1 - 6x^101 + 15x^202) *(1 + 6x + C(7, 2)x^2 + .... + C(n + 5, n)x^n + ...)
= C(245, 240) - 6C(144, 139) + 15C(43, 38 )
hi sekar, i cant understand how you get lowerlimit and upper limit. can you explain once again with clear approach.
Hi All,
I have a doubt in PnC..
when we have a word say : INDEPENDENCE while finding the number of arrangements of the same , why do we have to divide the total arrangements by the number of repetitions.
eg in this case : 12!/3!*2!*4!
here the number of way E can be arranged amongst itself is 4! , but y divide it by the total number of ways ?
:banghead:
Sorry for being 
Puys.I have opened a new thread of Verbal,which deals with odd word/pair out,antonym-synonym type questions.All are welcome 
:
The thread is http://www.pagalguy.com/discussions/odd-word-pair-out-and-words-relationship-for-mba-2011-25069070/2821305
Hi All,
I have a doubt in PnC..
when we have a word say : INDEPENDENCE while finding the number of arrangements of the same , why do we have to divide the total arrangements by the number of repetitions.
eg in this case : 12!/3!*2!*4!
here the number of way E can be arranged amongst itself is 4! , but y divide it by the total number of ways ?
:banghead:
Just look this ex, No of words formed with BOB=BOB,BBO,OBB=3->3!/2
No of words formes with BOT=BOT,BTO,TOB,TBO,OBT,OTB=6->3!=6
actually if letters are same they can be treated as same entity
In permutation arrangemnet is important
If you take "n" disssimilar items in linear permuation takes 'r" at a time withour repition you use nPr.
suppose if you want number of arrangement in N items of its,p is one kind and q is second kind.
you use formula N!/p!q!
Students score in each section is a, b, c, d, then
a + b + c + d = 90
=> C(93, 90) ways, but we have counted those cases also when a, b, c, d > 45
When a > 45, i.e., a = 46 + a'
=> a' + b + c + d = 44
=> C(47, 44) ways
Similarly for other variables
So, total number of ways = C(93, 90) - 4*C(47, 44) = 64096
hi ouy
i have a doubt on ur sol. for this prob. if a'=0 then u have to apply the form. of the partition theorem i.e when any of the partition can b empty
plese help me ....
Quote:
Originally Posted by chillfactor View Post
Students score in each section is a, b, c, d, then
a + b + c + d = 90
=> C(93, 90) ways, but we have counted those cases also when a, b, c, d > 45
When a > 45, i.e., a = 46 + a'
=> a' + b + c + d = 44
=> C(47, 44) ways
Similarly for other variables
So, total number of ways = C(93, 90) - 4*C(47, 44) = 64096
hi ouy
i have a doubt on ur sol. for this prob. if a'=0 then u have to apply the form. of the partition theorem i.e when any of the partition can b empty
plese help me ....
Question
Hi All,
1. In an 'akhada' m wrestlers stand around a circle. Each possible pair of persons not standing next to each other play a match with in the middle of the circle for 1 min and 30 secs. If the total time taken for all matches is 30 mins, then m is equal to??
2.Six boxes numbered 1 to 6 are arranged in a row.Each is to be filled by either a blue or a green ball such that no two adjacent boxes contain green colored balls. In how many ways can the boxes be filled with the balls?
Elaborate solution with proper explanation is required.
Thanks in advance.
Question
Hi All,
1. In an 'akhada' m wrestlers stand around a circle. Each possible pair of persons not standing next to each other play a match with in the middle of the circle for 1 min and 30 secs. If the total time taken for all matches is 30 mins, then m is equal to??
2.Six boxes numbered 1 to 6 are arranged in a row.Each is to be filled by either a blue or a green ball such that no two adjacent boxes contain green colored balls. In how many ways can the boxes be filled with the balls?
Elaborate solution with proper explanation is required.
Question
Hi All,
1. In an 'akhada' m wrestlers stand around a circle. Each possible pair of persons not standing next to each other play a match with in the middle of the circle for 1 min and 30 secs. If the total time taken for all matches is 30 mins, then m is equal to??
2.Six boxes numbered 1 to 6 are arranged in a row.Each is to be filled by either a blue or a green ball such that no two adjacent boxes contain green colored balls. In how many ways can the boxes be filled with the balls?
Elaborate solution with proper explanation is required.
My take on this set:
1. Each wrestler plays (m-3) matches.
So, m wrestlers will play m*(m-3)/2 matches.
... Dividing by 2 as one match is counted twice, once of each wrestler.
And each match takes 90 seconds.
So, number of matches = 30*60/90
So, m*(m-3)/2 = 30*60/90
=> m=8
2. Question recently answered on quant thread.
For 0 green balls, 1 way.
For 1 green ball, 6 ways.
For 2 green balls, 5C2 ways. ......
For 3 green balls, 4C3 ways. ......
For 4, 5 and 6 green balls, 0 ways.
So, totally 1+6+5C2+4C3 = 1+6+10+4 = 21 ways
Please help with this:
How many permutations of the word INTERMEDIATE can be formed such that:
(i)All vowels occupy even places.
(ii)Relative order of vowels does not change.
My approach:
(i)6 consonants are arranged in 6 odd places in 6! ways, and 6 vowels are arranged in 6 even places in 6!/(3!*2!) ways.
Answer = 6!*6!/(3!*2!) = 43200.
(ii)6 consonants are arranged in 6 places in 6! ways, and the vowels are shuffled around in their present places in 6!/(3!*2!) ways.
Answer = 43200.
OA is given to be 21600, in either case. Apparently, I am missing a 2! in a denominator somewhere, can someone point it out?
Please help with this:
How many permutations of the word INTERMEDIATE can be formed such that:
(i)All vowels occupy even places.
(ii)Relative order of vowels does not change.
My approach:
(i)6 consonants are arranged in 6 odd places in 6! ways, and 6 vowels are arranged in 6 even places in 6!/(3!*2!) ways.
Answer = 6!*6!/(3!*2!) = 43200.
(ii)6 consonants are arranged in 6 places in 6! ways, and the vowels are shuffled around in their present places in 6!/(3!*2!) ways.
Answer = 43200.
OA is given to be 21600, in either case. Apparently, I am missing a 2! in a denominator somewhere, can someone point it out?
Letters repeated
I-2
E-3
T-2
I m guessing you missed the "T" repeatition....
Letters repeated
I-2
E-3
T-2
I m guessing you missed the "T" repeatition....
:banghead:
Thanks...