puzzle

[saro_gn]

hi
pls any one help me find the answer in the following puzzle:

There are 10 steps. A person is starting from the bottom climbing to the top. A person can climb, at a time, one, two or three steps. In how many ways can he climb the steps and reach the top?

bye.,
saravana.

[mediator]

Well the answer is very simple..... U see there aare ten steps and three given step incremenets => single, double , triple. So the equation formed is :

3x+2y+1z = 10

Just count the number of sets x,y,z form. Thats the solution. Like x=3,y=0,x=1 is one set/way. So count the number of these ways !!!

[sakumar79]

There is a twist to the above solution presented by mediator... It will count steps of (2, 3, 2, 3) and (3,2,3,2) and (2,2,3,3), etc to be same since x=0, y=2,z=2 for all cases... But, in this situation, they should be counted as different... So, we are looking at permutations rather than combinations, but it is not as direct as that because of the restraint in the sum...

However, what this post is doing in a QnA thread is more puzzling

Arun

[hermit]

taking 1 step --> 1 way ;
taking 2 steps --> 1 way;
combination of 2&3 --> 4 ways;
combination of 1&3 --> 4 ways;
combination of 1,2 &3-->5!/5*2! + 3!

ways

[sakumar79]

Originally Posted by hermit

taking 1 step --> 1 way ;
taking 2 steps --> 1 way;
combination of 2&3 --> 4 ways;
combination of 1&3 --> 4 ways;
combination of 1,2 &3-->5!/5*2! + 3!

ways

1&2 combination has been left out above, and some mistakes are there...

Combination of 2 & 3: Here, no. of 3s has to be 2 (cannot be 0 as it will be all 2s which has been accounted, cannot be 1 or 3 as there will be odd number in total, anything above 3 will lead to total more than 10.). So, you have 2 2s and 2 3s in this combination set... (2,2,3,3), (2,3,2,3), (2,3,3,2), (3,2,2,3), (3,2,3,2), (3,3,2,2,) implies 6 combinations...

Combinations of 1 & 3: Here, things get a bit complicated...If we have x 1s and y 3s, x+y must be even, but x and y can both be odd... x and y must be greater than 0... y=1 => x=7, y=2 => x=4, and y=3 => x=1.
x=7, y=1 => 8 combinations
x=1, y=3 => 4 combinations
x=4, y=2 => (111133) (111313) (111331) (113113) (113131) (113311) (131113) (131131) (131311) (133111) (311113) (311131) (311311) (313111) (331111) 15 combinations...

Combinations of 1 & 2: Here again, the combinations will increase... no. of 1s must be 2, 4, 6 or 8. I will get back to this later...
2 1s, 4 2s
4 1s, 3 2s
6 1s, 2 2s
8 1s, 1 2(s)

And then we have combinations of 1,2 and 3, which I will look into later...

Arun

[mediator]

Hey that what i meant! taking x=1, count all possible positive values of y and z, then x=2 => y,z ?.....then x=3 =>y,z ? so and so....means all the sets formed is the answer!

[martian]

Mm.. Maths! Maths!

Wait, I've got a Puzzle too!

Lets suppose a Farmer owns 50 Cows... and has 9 rooms to lock them up! One day it starts to rain. He can lock any number of Cows in each room, but can do so only in Odd numbers... So how many Cows in each rooms at last?

[mediator]

Ur problem is same as above! It shud be how many ways the cows can be entered in the rooms with the given condition.
no. of cows in each room will vary in each step!

[martian]

Originally Posted by mediator

Ur problem is same as above! It shud be how many ways the cows can be entered in the rooms with the given condition.
no. of cows in each room will vary in each step!

So how many Cows in each room Mediator?

[mediator]

@Martian.....Bro .....Im not a supergenius! Have mercy! I gave up studying permutation and combinations a long time back.
But in ur case the eqaution will be similar as previous problem!
I dunno how to solve that equation now! If i cud...then i wud have posted the exactanswer too!