[rohit pandeyParticipant]

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

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  rohit pandeyQuote

[rohit pandeyParticipant]

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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[rohit pandeyParticipant]

and the same problem for a circle.

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[rohit pandeyParticipant]

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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[rohit pandeyParticipant]

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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[rohit pandeyParticipant]

answer is 2^6

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[rohit pandeyParticipant]

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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[rohit pandeyParticipant]

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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[rohit pandeyParticipant]

the answer is {6*7!/2!}/6^7

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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[rohit pandeyParticipant]

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}
this is a probable answer

where did you get this question?

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[rohit pandeyParticipant]

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

and add them up

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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[rohit pandeyParticipant]

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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[rohit pandeyParticipant]

no the budget line will not touch the axes.

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[rohit pandeyParticipant]

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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