Permutations & Combinations - Questions & Discussions

@NishantGupta23 said:

Please answer the following question:A and B pick up a card at random from a well schufle pack of cards,one after the another,replacing it every time till one gets a heart.If A beins the game then the probability of B ends the game is :1) 2/7 2)4/7 3)3/4 4)1/4 5)none of these

If A starts the game, and B ends it, that means whatever be the number of times they pick it up, A never got hearts and B got the heart at the end.

If it was just two pick ups then Probability of that is (A didn't get hearts) followed by (B got hearts)

P = (3/4)*(1/4)

If it took four chances, P = (3/4)*(3/4)*(3/4)*(1/4) .. which basically means that in the previous three chances no one got a heart.

so, this way it may go on till infinity...

So, final probability of these mutually exclusive events will be the summation of all these probabilities.
P(B winning) = (3/4)*(1/4) + (3/4)*(3/4)*(3/4)*(1/4) +.... infinity..
this is a GP where a = (3/4)*(1/4) and r = (3/4)*(3/4)

Solving this, we get P = 3/7

There are 4 identical oranges, 3 identical mangoes and 2 identical apples in the basket. the number of ways in whihc we can select one or more fruits from the basket is

Ans:60, 59, 57, 55, 56

Please help in solving this

In an examination , max marks for 3 papers is 50 each and for the fourth is 100 . Find the number of ways in which a student can score 60% aggregate marks .


A. 330850
B. 233551
C. 110551
D. 220800
Hello Puys,
I have a ppt related to similar and differnt concept in permutation combination.
we have four combnations (say ball and box case:
1 > similar ball similar boxes
2 > similar ball differnt boxes
3> different ball same boxes and
4> different ball different boxes.
In the attaced ppt the all the four cases are discussed.
I hope it will be helpful... :)

There are 25 points on a plane of which 7 are col-linear , how many quadrilaterals can be formed from these points?

a)5206
b)2603
c)13015
d)none of these

Funda of calculating the number of triangles is clear to me but not of calculating the number of quadilaterals.

@mayankdangi said: There are 25 points on a plane of which 7 are col-linear , how many quadrilaterals can be formed from these points?
a)5206
b)2603
c)13015
d)none of these
Funda of calculating the number of triangles is clear to me but not of calculating the number of quadilaterals.
Is it none of these ?

25C4 - 7C4 - 7C3*18C1 = 11985 ?
@YouMadFellow said:
Is it none of these ?
25C4 - 7C4 - 7C3*18C1 = 11985 ?
Yes u are right.
Can u please elaborate?
TIA
@mayankdangi said:
Yes u are right.
Can u please elaborate?
TIA

Im getting none of these , not sure though

Total -( all 4 points are collinear points + 3 points on collinear points&other; point is a non collinear point)

25C4-(7C4+ 7C3*18)

Solving gives none of these

Explained in detail in the arun sharma thread u had just posted ( not able to tag u now) some problem

Buddy posting in a single thread will be enough

@mayankdangi said:
Yes u are right.
Can u please elaborate?
TIA
quadrilateral can be formed by 4 points

Now the not possible cases are all four points lying in that collinear points and three out of four lying in the collinear points ..

so 25 C 4 - 7c4 - 7c3*18c1

:D

Apologize for posting repeatedly.
I still did nt get the part where 3 points on collinear points and point is a non collinear point is mentioned.
@mayankdangi said:
Apologize for posting repeatedly.
I still did nt get the part where 3 points on collinear points and point is a non collinear point is mentioned.
See when you select three points on a line, and another point outside somewhere in the space, how can you form a quadrilateral ? So, we need to remove those cases from the total cases.

PS: I hate "none of these" option
@YouMadFellow said:
See when you select three points on a line, and another point outside somewhere in the space, how can you form a quadrilateral ? So, we need to remove those cases from the total cases.
PS: I hate "none of these" option
Thank you @[562114:Budokai001], @[592846:YouMadFellow], @[169132:naga25french]

If all the words formed by letters "RAINBOW" are arranged in a dictionary form,then what is the position of the word RAINBOW in that dictionary?

@sidroy09 said: If all the words formed by letters "RAINBOW" are arranged in a dictionary form,then what is the position of the word RAINBOW in that dictionary?
arrange in ascending order ABINORW

each alphabet will have 6! words

since ABINO are before R , we have 5*6! words

next for R , A is the least so its correctly placed . B should occur next .. but we have I .. so 24 words are taken ..

And if I comes first , next should be B .. so we need to count another 3! ..

Rest all words are in correct position

so intotal 3600 + 24 + 4 = 3630 .. so RAINBOW would be 3631th word

naga sir this is for a word where there is no repetition suppose for the word

MONALISA
which word would be monalisa how will it differ
@naga25french said:
arrange in ascending order ABINORW
each alphabet will have 6! words
since ABINO are before R , we have 5*6! words
next for R , A is the least so its correctly placed . B should occur next .. but we have I .. so 24 words are taken ..
And if I comes first , next should be B .. so we need to count another 3! ..
Rest all words are in correct position
so intotal 3600 + 24 + 4 = 3630 .. so RAINBOW would be 3631th word

Thnx 😃 ...It was pretty easy infact 😛 .. Dont know where my mind went that time 😛 ...

And there is one Typo.... its 3600+24+6=3630 :)

@phoenix2011 said:naga sir this is for a word where there is no repetition suppose for the word
MONALISA
which word would be monalisa how will it differ

@[453074:phoenix2011] : Is the Answer 2535?

@[169132:naga25french] : Please verify my answer.

@naga25french said:
arrange in ascending order ABINORWeach alphabet will have 6! wordssince ABINO are before R , we have 5*6! wordsnext for R , A is the least so its correctly placed . B should occur next .. but we have I .. so 24 words are taken ..And if I comes first , next should be B .. so we need to count another 3! ..Rest all words are in correct positionso intotal 3600 + 24 + 4 = 3630 .. so RAINBOW would be 3631th word
But
@chillfactor said: Yup my mistake. Thanks for the correction.Lets say that a is the marks lost in one section, b in another, c is third and d in fourth.So, (45 - a) + (45 - b) + (45 - c) + (45 - d) 90 a + b + c + d 90So, C(94, 4) ways, but we have counted those cases also when a, b, c, d > 45When a > 45, i.e., a = 46 + a' => a' + b + c + d 44 => C(48, 4) waysSimilarly for other variablesSo, total number of ways = C(94, 4) - 4*C(48, 4)
Sir:

93C3 hoga and 47C3 hoga which will give 64906 ways.

@Estallar12 said:
Sir:
93C3 hoga and 47C3 hoga which will give 64906 ways.
Question says that 90 is qualifying marks and in how many ways one can qualify, so he can get more than 90 also.

thats why greater than equal to sign is there.