# rohit pandey

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Viewing 14 posts - 1 through 14 (of 14 total)
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• in reply to: permutation and combination #554

rohit pandey
Participant

this is if first choice is g1 which has ten ways now for b1 also there willbe 10 ways.

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• in reply to: de groot #552

rohit pandey
Participant

got the above questions
another question
n people are seated in random manner in a row containing 2n seats ,what is probability that no two people will occupy adjacent seats.

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• in reply to: de groot #551

rohit pandey
Participant

and the same problem for a circle.

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• in reply to: probability #543

rohit pandey
Participant

if we take m=1 and n=5 it is not divisble by 5 so we cant generalize m and nas odd will always yield numbers divisible by 5 .

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• in reply to: permutation and combination #542

rohit pandey
Participant

shruti the answer for q2 is indeed 20 but i will have to explain it in class.hint -try forming a tree its really big.

and for q1 if you take and sum the different permutations of 7 numbers form set you will find the sum of 4 options is divisible by 9 hence 4*7!.

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• in reply to: permutation and combination #520

rohit pandey
Participant

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• in reply to: probability #515

rohit pandey
Participant

just one correction p(a)=60c10/90c10
and p(a∩b)=30c10/90c10

so p(aubuc)=3*{60C10-30C10}/90C10

sorry had misplaced the values as i was sleepy

and p(aubuc) covers the possibility of two balls not being selected.

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• in reply to: probability #507

rohit pandey
Participant

take three events
A={NO RED BALL}
B={NO BLUE BALL}
C={NO WHITE BALL}
SO WE NEED TO FIND P{AUBUC}=?

SO P(A)=30C10/90C10 P(B) AND P(C) WILL BE SAME
NOW P(A∩B)=60C10/90C10 P(B∩c) and p(A∩C) WIL BE SAME

NOW P(A∩B∩C)=0 {CANT HAPPEN}
SO P(AUBUC}=3*{30C10-60C10}/90C10

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• in reply to: probability #504

rohit pandey
Participant

realised the above answer is wrong
first there are 7! ways for dice to get numbers since two numbers are same so divide by 2! and since there are six numbers that will be repeated there are 6 cases so multiply it by 6

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• in reply to: permutation and combination #502

rohit pandey
Participant

for q 1

even digits are 0 , 2 , 4 ,6 ,8

so probable 4 digits are 1.0246=0+2+4+6=12
2.2468=2+4+6+8=20
3.4680=4+6+8+0=18
4.6802=6+8+0+2=16
5.8024=8+0+2+4=14

only option 1 and option 3 is divisible by 3

so permutation of option1=4!-3!
( we do -3! factorial because we need 4 digit numbers and thus we fix 0 at fourth place and form permutations and subtract it from total permutations}
and
permutation of option2=4!-3! {same thing again}

where did you get this question?

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• in reply to: permutation and combination #501

rohit pandey
Participant

for q2
at least 2 digits same means
we take cases for
1.2 same digits
2.3 same digits
3.4 same digits

for case 1-5*4*3*2/2!
for case 2 – 5*4*3*2/3!
for case 3 – 5*4*3*2/4!
now add all of them up
this is a probable solution

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• in reply to: probability #500

rohit pandey
Participant

there are 7 different dices _ _ _ _ _ _ _
for 1st dice it can take any 6 numbers but other 6 dices will have permutation of 6*5*4*3*2*1*6
total no of events =6^7

so probability=6*5*4*3*2*1*6/6^7

maybe right maybe wrong

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• in reply to: Varian workbook q.2.11 #496

rohit pandey
Participant

no the budget line will not touch the axes.

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• in reply to: Varian workbook q.2.11 #489

rohit pandey
Participant

dont confuse yourself with variables at first just follow the question and calculate coordinates for 5000\$ on boring business magazine and trendy consumer magazine.

the question is about plotting points on graph and the calculate slope using y-y1=m(x-x1)
where m is slope m=y2-y1/x2-x1.
lawyer mba
5000\$ on business mag(3000 , 10000)(x1,y1)
5000\$ on consumer mag(5000 , 6000 )(x2,y2)

so y-10000/x-3000=-4000/2000

so the budget line is y+2x=16000
where y=M and x=L

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