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Hi,
I have been part of MBA entrance tests season twice, I have shared my experiences @ The CAT~Perience ~~Writing @ Will~~ , hope this will be useful
Also , my GD/PI experiences are @ The CAT~Perience ? Level II ( GD PI Hurdle) ~~Writing @ Will~~
Merry Reading,
Nitin
a)
as all 4 balls are same so if we arrnage them among them selves then onli one way , let me explain it further
lets take 4 rs1 coins to b given to 4 kids n each kid shd b given 1 rs exactly , there is only one way u can give , it doesnt matter which coin u give to which boy , end result will b same as all coins are identical .
we can arrange these 4 IDENTICAL balls in 5 DIFF boxs in 5x4x3x2
u can derive this very simply lik this : 1st red ball will hv 5 boxes to choose , or in other words we can put 1st red ball in any of 5 diff boxes, so wen1st red ball takes 1 box then 2nd red ball can b put in ny of remaining 4 diff bxes , simlilarly 3 rd ball can b put in ny of remaining 3 boxes , 4th ball in ny of remaining 2 boxes
so we get 5x4x3x2
we put x( multiplication sign ) in between 5 4 3 n 2 bc we r using word AND
we will read e solution lik : we can put 1st red ball in any of 5 diff boxes AND 2nd red ball can b put in ny of remaining 4 diff bxes AND 3 rd ball can b put in ny of remaining 3 boxes AND 4th ball in ny of remaining 2 boxes
so ans is 5x4x3x2x1( this 1 is for 1 way of arranging balls )
b)
4 diff balls n 5 same boxes , so in this case boxes wont matter , balls will
we can choose 4 identical boxes out of 5 identical boxes in 5c4 ( if there were only 4 identical boxes n we hv to choose 4 out of them then we whd hv had only one way of doing it , we can prove it by formula also 4c4 = 1 .....) tat is equal to 5 AND we can arrange 4 DIFF balls in 4! ways
as we used AND in above line , e solution wil b like
5 x 4!
c)
all balls same , so ball can b arranged in 1 way only n as all boxes same so we can choose 4 identical boxes out of 5 identical boxes in 5c4=5
so ans is
1x5=5
i dont noe abt formula u mentioned above , our faculty told us to take step by step approach or else this topic can b highly confusing
i guess above answer for question a is wrong. since boxes are different the number of ways in which 4 similar balls can be kept in 5 different boxes will be 5c4 i.e. 5. suppose the boxes are numbered as 1, 2,3,4,5 so 4 balls can be kept in either boxes 1,2,3,4 or 1,3,4,5 similarly 3 more options will be there. Hope I am right.
Ques1. There is a set of no. a1
Helloooo puys!
Please help me with the following question:
4 couples, i.e. 4 women and 4 men are to be seated around a round table in such a way that no men sit together and no woman sits next to her husband.
Find the number of arrangements.
Hope the question is clear. 😃
Helloooo puys!
Please help me with the following question:
4 couples, i.e. 4 women and 4 men are to be seated around a round table in such a way that no men sit together and no woman sits next to her husband.
Find the number of arrangements.
Hope the question is clear. :)
12 ways??? I suck badly in PnC..so giving it a try:shocked:
Helloooo puys!
Please help me with the following question:
4 couples, i.e. 4 women and 4 men are to be seated around a round table in such a way that no men sit together and no woman sits next to her husband.
Find the number of arrangements.
Hope the question is clear. :)
First of all arrange the men around the round table so that they occupy alternate positions
So 4 men can be arranged in 3!=6 ways around a round table.
Now the remaining 4 women will be arranged in the remaining four places of the round table. This is equivalent to arranging 4 women in a row=4!
Total arrangements= 4!.3!
Now since no woman should sit beside her husband so we can subtract the cases when the couple sits together.
No. of arrangements that can be made such that husband sits with his wife= 3!
Hence total arrangements= 144-6=138
Please confirm the answer... Not very confident with the solution.:shocked:
my take :
As above arranging the alternate men and women can be done in 3! x 4! ways = 144
now the no of ways that a couple sits together is 3! x 2 as the wife can sit either on the right or the left side of the husband.
hence ans = 144 -12 = 132
i hope its right ;)
my take :
As above arranging the alternate men and women can be done in 3! x 4! ways = 144
now the no of ways that a couple sits together is 3! x 2 as the wife can sit either on the right or the left side of the husband.
hence ans = 144 -12 = 132
i hope its right ;)
Haahahaha,look at my answer
Helloooo puys!
Please help me with the following question:
4 couples, i.e. 4 women and 4 men are to be seated around a round table in such a way that no men sit together and no woman sits next to her husband.
Find the number of arrangements.
Hope the question is clear. :)
I think its 12.
We can arrange the 4 men in 3! = 6 ways around the circle.
Now, there are just two of arranging the 4 women between the 4 men such that no couple sits together (you can try it manually, it won't take much time)
So, total 6*2 = 12 ways
I think its 12.
We can arrange the 4 men in 3! = 6 ways around the circle.
Now, there are just two of arranging the 4 women between the 4 men such that no couple sits together (you can try it manually, it won't take much time)
So, total 6*2 = 12 ways
What the eff!!!!!i have got the same answer as chillfactor
Is this for real????
I think its 12.
We can arrange the 4 men in 3! = 6 ways around the circle.
Now, there are just two of arranging the 4 women between the 4 men such that no couple sits together (you can try it manually, it won't take much time)
So, total 6*2 = 12 ways
@ chillfactor
Sir jee ....If u will then kindly shed some light on why is our method wrong.
Where exactly did we screw up. Thanks
@spectra
kyuun bhai - why can't you be right 😉 aakhir you are the quants guy before you became the verbal guy
@spooky3 you have written that no. of ways in which the men and women can be arranged alternately is 3! x 4! ...may i know how you came to this? is it because you arranged the no. of men first in a circle (4-1)! and then filled up the empty spaces with the no. of women i.e. 4 women remaining so 4!?
p.s. i suck at quant ...just getting my doubts cleared.
Ques1. There is a set of no. a1<............. can="can" how="how" in="In" many="many" select="select" u="u" ways="ways">
uhh Q1. 10c3 expl: from 10 numbers you choose any 3 , one of the ways they can be arranged is the ascending order and Combination just gives one way of selecting so..... what is the correct ans?
Q2. Total no. of ways in which yo can select atleast one from n items is 2^n - 1... so in ur case 2^(2n+1) - 1 = 63 should be the working , however n is coming as a fraction so i would urge you to check the question again...( or please post the working of the solution if you get it so i can correct my solution...:-) )
Q3. total no. of ways in which section A can be attempted is ( 3c1+3c2+3c3) similarly section B can be attempted in (4c2+4c3+4c4) so if you multiply the 2 i think the ans should come.
@spooky3 you have written that no. of ways in which the men and women can be arranged alternately is 3! x 4! ...may i know how you came to this? is it because you arranged the no. of men first in a circle (4-1)! and then filled up the empty spaces with the no. of women i.e. 4 women remaining so 4!?
p.s. i suck at quant ...just getting my doubts cleared.
exactly right man ....... i did it like that.
thus proved : you do not suck in quants ;)
though my approach is not in congruency with chillfactor and i would like to know the place where ( in case ) i went wrong.
spectramind07 SaysWhat the eff!!!!!i have got the same answer as chillfactorIs this for real????
Seems you are a highly talented person, one can't hide his/her talent for a long time, one day or the other, it will be known to everyone. :cheers:
@ chillfactor
Sir jee ....If u will then kindly shed some light on why is our method wrong.
Where exactly did we screw up. Thanks
@spectra
kyuun bhai - why can't you be right 😉 aakhir you are the quants guy before you became the verbal guy :cheerio:
When you are subtracting the 12 cases, you are not considering which couple you are choosing and even others can be permuted among themselves.
Even if we consider this, then there will be over counting as it might be possible that 2 or 3 or 4 couples are sitting together. So I think it will get complicated if we proceed this way.
I hope it makes sense. (I'm very bad at explaining stuff)
I think its 12.
We can arrange the 4 men in 3! = 6 ways around the circle.
Now, there are just two of arranging the 4 women between the 4 men such that no couple sits together (you can try it manually, it won't take much time)
So, total 6*2 = 12 ways
This question came in my last CL mock. The answers arent out yet :(
But I also got 12 as my answer..by applying the same concept.
Couldn't wait for the answers to come out..! So posted early :)
Will post the answer on the 13th of this month.. 😃 Bear with me 😃
This question came in my last CL mock. The answers arent out yet :(
But I also got 12 as my answer..by applying the same concept.
Couldn't wait for the answers to come out..! So posted early :)
Will post the answer on the 13th of this month.. 😃 Bear with me :)
I also think the answer is 12. Got it wrong before...
Originally Posted by Mugs
Can anybody explain this problem from CL sheet
There are 7 distinct coins which r 2 b distributed randomly into 5 distinct pots in a row. Any pot can receive any no. of coins. All coins need to b distributed.
In how many ways can these coins b distributed so that no pot is empty?
Can some1 explain using the grouping approach
another method to solve this prob.
each pot has to get one coin.. hence we can arrange one coin in each pot in 7P5 ways.. the remaining 2 coins can be placed in the 5 pots in 5P2 ways.. hence the total no of ways is 7P5*5P2= 16800
There are 7 distinct coins which r 2 b distributed randomly into 5 distinct pots in a row. Any pot can receive any no. of coins. All coins need to b distributed.
In how many ways can these coins b distributed so that no pot is empty?
Total number of ways = 5^7
but here we have counted those cases also when one of the pot doesn't have any coin, no of such cases = 5*4^7 (as now we have to fill only 4 pots, and we can chose the empty pot in 5 ways)
Here also, we have counted those cases when two pots are empty.
Number of such cases = 10*3^7 (we can chose the two empty pots in C(5, 2) ways)
Here also, we have counted those cases when three pots are empty.
Number of such cases = 10*2^7 (we can chose the three empty pots in C(5, 3) ways)
Here also, we have counted those cases when four pots are empty.
Number of such cases = 5*1^7 (we can chose the four empty pots in C(5, 4) ways)
=> Total = 5^7 - 5*4^7 + 10*3^7 - 10*2^7 + 5 = 16800
another method to solve this prob.
each pot has to get one coin.. hence we can arrange one coin in each pot in 7P5 ways.. the remaining 2 coins can be placed in the 5 pots in 5P2 ways.. hence the total no of ways is 7P5*5P2= 16800
I don't think we can use this method. Even P(7, 5)*P(5, 2) is not 16800