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amistre64 Group TitleBest ResponseYou've already chosen the best response.0
i can try ...
 2 years ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
hi thx for cming its urgent
 2 years ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
1.there are 3 sections in a question paper ,each containing 5 qns .A candidate has to solve any 5 qns ,choosing atleast one qn from each section .in hw many ways can he make a chice?
 2 years ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
2.There are K intermediate stations on a railway line from one terminus to anotherin hw many ways can a train stop at 3 of these intermediate stations if no 2 of these stopping stations are to be consecutive??
 2 years ago

LoveYou*69 Group TitleBest ResponseYou've already chosen the best response.1
I dont know... Sorry
 2 years ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
pls help its urgent
 2 years ago

darthsid Group TitleBest ResponseYou've already chosen the best response.1
Here's my attempt for 1: Since the candidate has to choose at least one question from each section, the ways of choosing the first question are 5, similarly the second question and third question can be chosen in 5 ways each. So, total ways in which first 3 questions can be chosen is 5x5x5 = 125. Now, there are 12 remaining questions, and he has to choose 2 of them. So, the remaining questions can be chosen in 12C2 ways. Thus, total is 125 x 12C2 = 125 x 66 ways
 2 years ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
pls help with othr one
 2 years ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
3.There are n points in a plane of which no 3 are in a straight line except m which are all in a straight line .Find the no: of (A)different straight lines (b)different triangles (c)different quadrilaterals that can be formed with the given points as vertices.
 2 years ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
4.find the sum of all 5 digit nos: formed by digits 1,3,5,7,9 when (!)no digit is being repeated (2)repetition is allowed
 2 years ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
next one james
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.2
Aravind, much better you post one question at a time and digest the answers to one before you move on to the other. If you understand this answer, you will be better able to tackle the next one on your own.
 2 years ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
james i dont hav time its 12 :30 here pls help me
 2 years ago

AravindG Group TitleBest ResponseYou've already chosen the best response.1
its urgent :((
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
they shut OS down and then i went to class
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
i was gonna do darths; 5^3 * 12P2
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
12P2 simply since the order in which you choose them is considered a different choice
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
i think james has what, 60*60*5 = 3600*5 = 18000 5^3 = 125*120 would be my results, i hope tey match lol
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
16500 was mine ..... prolly wrong
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.2
If there are three sections, X, Y, Z, there are a number of different ways to choose 5 questions X Y Z 2 2 1 2 1 2 1 2 2 3 1 1 1 3 1 1 1 3 The number for the first three is \[ 3{5 \choose 2}{5 \choose 2}{5 \choose 1} = 3(10)(10)(5) = 1,500 \] and the number for the second three is \[ 3{5 \choose 3}{5 \choose 1}{5 \choose 1} = 3(10)(5)(5) = 750 \] giving a total of \( 2,250 \). I think the problem with DarthSid's method is there is double (or more) counting. For example, call the questions x1,x2,x3,x4,x5 y1,y2,...,y5 z1,...,z5. Then one of the options under DS's algorithm is 1. Select one from each of x,y,z: let's take x1,y1,z1. 2. Now two of the remaining: let's take x2,y2 Hence the selection is x1,x2,y1,y2,z1 But we can generate that same selection by 1. taking x2,y2,z1 2. then x1,y1 or 1. x1,y2,z1 2. then x2,y1
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
my thought process was, and i hold no accuracy for it, was sections 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 there are 5^3 ways to satisfy 1 question from each section to get 3 questions that leaves us with 12 questions to arbitrarily pick 2 more questions from in a random order. Hence, 12*11 more results .... but James, as always, looks much more convincing ;)
 2 years ago

JamesJ Group TitleBest ResponseYou've already chosen the best response.2
What you've laid out is DS's picking algorithm as well. The trouble is that algorithm repeats selections. Took me a minute to figure it out though, because at first blush it looks good.
 2 years ago
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