@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
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 .
There are 25 points on a plane of which 7 are col-linear , how many quadrilaterals can be formed from these points?
@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)5206b)2603c)13015d)none of theseFunda of calculating the number of triangles is clear to me but not of calculating the number of quadilaterals.
@YouMadFellow said:Is it none of these ?25C4 - 7C4 - 7C3*18C1 = 11985 ?
@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
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
@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.

@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
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?
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
@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
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 wordMONALISAwhich 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
@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)
@Estallar12 said:Sir:93C3 hoga and 47C3 hoga which will give 64906 ways.