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2) the no of ways of arranging 7 persons (a,b,c,d among them) in a row so that a,b,c,d are always in the order a-b-c-d (not necessarily together) will be

- 13Comments
- total arrangment is 7! , now in this 7! arrangmnt a,b,c,.... 3d.
- The arrangement has to be of the form PaQbRcSdT (the rest.... 3d.

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Hemante Singhal
@hemante
72

total arrangment is 7! , now in this 7! arrangmnt a,b,c,d will be arranged in 4! (not necessarily together) , but u need only one of those 4! , so 7!/4! @webby

Saket Binani
@cat2013aspirant
26

The arrangement has to be of the form PaQbRcSdT (the rest 3 people can take places in P,Q,R,S,T)

where P + Q + R + S + T =3 (P,Q,R,S,T has non negative integers)

So no. of ways = (n+s-1) C (n-1) where (n is 5 and s is 3) = 35.

Now in this remaining 3 people can be arranged in 3! ways = 6 .

Total ways = 35*6 =210..

1) A bag contains unlimited no of balls of four distinct colours such the balls of same colour are indistinguishable. The no of diff ways of slecting 12 balls frm the bag will be

Plz explain

- 5Comments
- yes!! plz explain. 4d.
- a+b+c+d=12...a,b,c.d are no. of balls of different colour.... 4d.

Sumit Kumar
@sumit6361
16

a+b+c+d=12...a,b,c.d are no. of balls of different colours...so 15c3

how many negative integer sol exists for x^2-y^2 =400?

If x is a real no then what is the no of solutions for the eq:

((x^4-16)(x^2-4))^1/2?

the set of all real nos x for which x^2-|x+2|+x >0

- 1 Like 20Comments
- @saphira1 I think ur 3rd answer is wrong it can not be -i.... 29 Jul.
- i think the answer for third one should be .. from 2 to i.... 29 Jul.

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swati garg
@saphira1
37

Answer1 - 4 ways (-101, -99), (-52,-48) (-29, -21) and (-25, -15).

Answer2 - infinite solutions.

Answer 3 (-inf, -rt2) (rt2, inf)

nitin nandan
@nitinknandan
3

@saphira1 exactly my point, if u see the number of positive integral solutions will also be the same

swati garg
@saphira1
37

yeah. That's how I got it too. Only takeaway from the word "negative" is that |x|>|y|.

Anirban Kar
@Anirban_kar
14

Yes, @swati The total answers will be 8 only. We just have to take the even factors that make up 400 only. Total factors = 15, of which (5, 80), (25,16), & (1,400) will be taken off. So, it seems that the answer is 15 -6 = 9. But, again (20,20) will have only 1 solution instead of 2. So, 8 is the final answer. Thanks for the tips.

mayank asd
@mack01
50

@saphira1 I think ur 3rd answer is wrong it can not be -inf to -rt2?

_rininir_ _dev_
@asaktidev
1

i think the answer for third one should be .. from 2 to infinity

Here are few ques...

A speaks truth in 60% cases and B in 70% cases. Wat is the prob that they will do the same thng while describing a single event. If they both lie, they tell the same thing?

A nd B toss coin alternatively till one of them gets a head. If A starts find the prob that B will win the game.

the prob of picking a spade or an ace not of spade from a pack of 52 cards is?

- 9Comments
- third one is 4/13..plz explain other two also 2 & 3. 27 Jul.
- @webby the second one is P(B winning)= P(A gets tails)*.... 28 Jul.

alok k
@allok
2

@webby the second one is

P(B winning)= P(A gets tails)*P(B gets head) + P(A gets tails)*P(B gets tails)*P(A gets tails)*P(B gets head) +P(A gets tails)*P(B gets tails)*P(A gets tails)*P(B gets tails)*P(A gets tails)*P(B gets head) +. so on upto infinity

the probability of P(A gets tails)= P(B gets head) = P(B gets tails) = 1/2

So P(B winning) is a GP with infinite terms with first term a = 1/4 and r= 1/4

So P(B winning) =a/(1-r) = 1/4/(1-1/4)=1/3

For the 3rd one in a pack of cards there are 13 spades + 3 ace except the ace of spades

So we have probability =16/52 =4/13