CAT4MBA

Hi,

I am very weak in Permutation and combination. So all chances that the solution may go wrong. Guys help me clear my understanding. 

 1. The possible number of ways in which he can take the steps are: 2*6 + 1. Now, the single step he can take in six ways, i.e. between any two "2 steps", first step or the last step. So the number of 2 steps ways are: 6.

No. of one step ways are: 1.

Again he can mix the 2 steps and the I step. So, possible cases:

             1*2 + 11"1 = 12 ways

             2*2 + 9*1 =   10C2 = 45 ways

             3*2 + 7*1 = 8C3 = 56 ways

             4*2 + 5*1 = 6C4 = 15 ways

             5*2 + 3*1 = 6C3 ways= 20 ways.

               So, the total becomes: 6+1+12+45+56+15+20 = 154 ways.

 O don't thik the solution is proper. Even then I put it because I want u frnds to correct my understanding and tell me where I am going wrong. Please help me.

 The future belongs to those who beleive in the beauty of their dreams.

Part - 1
        It is given that there are 13 steps - and sachin can make either 1 step ar 2 step at time.
        Let x be number of 1 step and y be number of 2 steps the 1x+2y=13 where x and y are integers.
        This comprises of the following set S whose element are = {(1,6),(3,5),(5,4),(7,3),(9,2),(11,1)}

        The number of ways of taking 1 one step and 6 two step =  13!/(1!12!) = 13 = 13C1
        The number of ways of taking 3 one step and 5 two steps =  13!/(3!10!) = 286 = 13C3
        The number of ways of taking 5 one step and 4 two steps =  13!/(5!8!) = 1287 = 13C5
        The number of ways of taking 7 one step and 3 two steps =  13!/(7!6!) = 1716 = 13C7
        The number of ways of taking 9 one step and 2 two steps = 13!/(9!4!)= 715 = 13C9
        The number of ways of taking 11 one step and 1 two steps = 13!/(11!2!) = 78 = 13C11

        So total ways = 13+286+1287+1716+715+78 = 4095 Ans.
        (1+x)^13 for x =1 is 13C0+13C1+13C2+...+13C13 = 2^13
        (1+x)^13 for x = -1 is 13C0 - 13C1+13C2-...-13C13 = 0 => 13C0+13C2+13C4+...+13C12 = 13C1+13C3+...+13C13

         Thus it can be seen from above 13C1+13C3+...+13C11 = 2^13/2 -1 as 13C13 =1 => 4096 -1 = 4095 Ans.
   

Part -2
        It is given that there are 14 steps - and sachin can make either 1 step ar 2 step at time.
        Let x be number of 1 step and y be number of 2 steps the 1x+2y=14 where x and y are integers.
        This comprises of the following set S whose element are = {(0,7),(2,6),(4,5),(6,4),(8,3),(10,2),(12,1),(14,0)}
        So total ways = { 14C0 +14C2+14C4+14C6+14C8+14C10+14C12+14C14 } = 2^14/2 =2^13 = 8192 Ans.

Part - 3
        As the stair is moving up with velocity of lets say y where y = 1 step in 5 seconds.
       Total time taken on standstill stair = 13 /(speed = 2 steps in 5 sec ) = 65/2 = 32.5 sec
       When stair is moving up the relative speed = 3 step / 5 sec total time =  13*5/(3) =65/3=21.66 sec
       Time taken is less by 10.84 sec.

Part -4
       Speed of person = twice the sachin = 4 steps in 5 sec relative speed = 3 step in 5 second = 21.66 sec

Part -5
       Total time taken in down journey = 65 sec thus net time = 21.66+65 = 86.66 sec
      

 Ans to part-1 and 2 earlier submitted was incorrect as  I tested this by induction and it failed. The correct
       solution for part-1 for set group =  {(1,6),(3,5),(5,4),(7,3),(9,2),(11,1),(13,0)}
       Total ways = 7C1+8C3+9C5+11C9+12C11+13C13 = 7+56+1134+55+12+1=1265
       And for part-2 is set group={(0,7),(2,6),(4,5),(6,4),(8,3),(10,2),(12,1),(14,0)}
       Total ways = 7C0+8C2+9C4+10C6+11C8+12C10+13C12+14C14= 1+ 24+1134+210+165+66+13+1= 1614
  

As I wrote earlier I have not solved these questions on my own so cant tell what is the answer

If the stair case starts moving up at half the speed of sachin who takes 2 steps in 5 seconds then
3. How much more time would sachin take in moving up the stair case compared to the stand still stair case (Calculate for 13 step stair case) 

Sol.

speed of sachin = 2 steps in 5 sec

speed of stair =  1steps in 5 sec

speed of sachin (w.r.t. stair) = 3 step in 5 sec

total time = 65/3 sec

if sachin not moving, then time = 65 sec

diff = (65 - 65/3) sec = 130/3 sec  

4. How much time is required for one person who moves at twice the speed of sachin in moving down the stair case?

speed of man = 4steps / 5sec

speed of stair = 1step/ 5sec

speed of man (w.r.t. stair) = 3steps/5sec

total time = 65/3sec  

5. How much time sachin would take in moving once up and down in the moving stair case of 13 steps?

Sol.

moving up

speed of sachin (w.r.t stair) = 3steps/5sec

time = 65/3 sec

moving down

speed of sachin (w.r.t stair) = 1steps/5sec

time = 65 sec

total time = 195/3 sec

Hi, Now lets assume that sachin wants to go up the ladder by taking either one,two or three steps and number
      of steps be 13 then number of ways will be = 1299 way
      Here x+2y+3z = 13
      Consider the table below for values of x,y,z respectively
      | 1 step | 2 step | 3 step  |
      |    13    |    0     |    0      |     Number of ways =  13!/(13!)(0!)(0!) = 1
      |    11    |    1     |    0      |     Number of ways = 12!/(11!)(1!)(0!)  = 12
      |    10    |    0     |    1      |     Number of ways = 11!/(10!)(0!)(1!) = 11
      |     9     |    2     |    0      |     Number of ways = 11!/(9!)(2!) = 11x5 = 55
      |     8     |    1     |    1      |     Number of ways = 10!/(8!)(1!)(1!) = 90
      |     7     |    3     |    0      |     Number of ways = 10!/(7!)(3!) = 10x9x8/6= 120
      |     7     |    0     |    2      |     Number of ways = 9!/(7!)(2!) = 36
      |     6     |    2     |    1      |     Number of ways = 9!/(6!)(2!)(1!) = 9x8x7/2= 252
      |     5     |    1     |    2      |     Number of ways = 8!/(5!)(1!)(2!) = 168
      |     4     |    0     |    3      |     Number of ways = 7!/(4!)(3!)(0!) = 35
      |     3     |    2     |    2      |     Number of ways = 7!/(3!)(2!)(2!) = 210
      |     3     |    5     |    0      |     Number of ways = 8!/(5!)(3!)(0!) = 56
      |     2     |    4     |    1      |     Number of ways = 7!/(2!)(4!)(1!) = 105
      |     2     |    1     |    3      |     Number of ways = 6!/(2!)(1!)(3!) = 60
      |     1     |    6     |    0      |     Number of ways = 7!/(6!)(1!)(0!) = 7
      |     1     |    3     |    2      |     Number of ways = 6!/(1!)(3!)(2!) = 6x5x2=60
      |     1     |    0     |    4      |     Number of ways = 5!/(1!)(0!)(4!) = 5
      |     0     |    5     |    1      |     Number of ways = 6!/5! = 6
      |     0     |    2     |    3      |     Number of ways = 5!/(3!)(2!) = 5x2 =10
    Thus total number of cases = 1+12+11+55+90+120+36+252+168+35+210+56+105+60+7+60+5+6+10 = 1299 ways