Question

permutation help

Answer 1

i can try ...

Answer 2

hi thx for cming its urgent

Answer 3

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?

Answer 4

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??

Answer 5

I dont know... Sorry

Answer 6

:(

Answer 7

pls help its urgent

Answer 8

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

Answer 9

pls help with othr one

Answer 10

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.

Answer 11

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

Answer 12

k

Answer 13

next one james

Answer 14

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.

Answer 15

james i dont hav time its 12 :30 here pls help me

Answer 16

james?

Answer 17

its urgent :((

Answer 18

they shut OS down and then i went to class

Answer 19

i was gonna do darths; 5^3 * 12P2

Answer 20

12P2 simply since the order in which you choose them is considered a different choice

Answer 21

i think james has what, 60*60*5 = 3600*5 = 18000

5^3 = 125*120 would be my results, i hope tey match lol

Answer 22

16500 was mine ..... prolly wrong

Answer 23

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

Answer 24

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 ;)

Answer 25

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.