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AravindG
permutation help
hi thx for cming its urgent
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.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??
I dont know... Sorry
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
pls help with othr one
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.
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
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.
james i dont hav time its 12 :30 here pls help me
they shut OS down and then i went to class
i was gonna do darths; 5^3 * 12P2
12P2 simply since the order in which you choose them is considered a different choice
i think james has what, 60*60*5 = 3600*5 = 18000 5^3 = 125*120 would be my results, i hope tey match lol
16500 was mine ..... prolly wrong
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
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 ;)
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.