One-can-only-move-to-the-right-or-downwards-on-the-4-by-6-point-lattice-shown-How-many-paths-from-to-are-there-

Question Number 3273 by Yozzi last updated on 09/Dec/15

* * * * O n e c a n o n l y m o v e t o t h e

* * * * r i g h t o r d o w n w a r d s o n t h e

* * * * 4 b y 6 p o i n t l a t t i c e s h o w n .

* * * * H o w m a n y p a t h s f r o m * t o

* * * * * a r e t h e r e ?

* * * *

Answered by Filup last updated on 09/Dec/15

Assume you move *

You must move :

Down 5 times

Right 3 times

You can move in any c o m b i n a t i o n

just so long as all moves are made

∴ it is the permutation^{5} P_{3} = 60

Commented by Filup last updated on 09/Dec/15

I hope this is correct!

Answered by prakash jain last updated on 09/Dec/15

x_{1} d_{1} x_{2} d_{2} x_{3} d_{3} x_{4} d_{4} x_{5} d_{5} x_{6}

x n e e d s t o b e f i l l e d w i t h r_{1} r_{2} r_{3}

r_{1} r_{2} r_{3} t o g e t h e r 6 c h o i c e s

r_{1} r_{2} a n d r_{3} 5 + 4 + 3 + 2 + 1 = 15

(r_{1} r_{2} at x_{1} leaves 5 possibilities for r_{3})

r_{1} a n d r_{2} r_{3} 5 + 4 + 3 + 2 + 1 = 15

(s a m e a s a b o v e)

r_{1} r_{2} r_{3}

r_{1} a t x_{1} x_{1} x_{2} x_{3} x_{1} x_{2} x_{4} x_{1} x_{2} x_{5} x_{1} x_{2} x_{6}

x_{1} x_{3} x_{4} x_{1} x_{3} x_{5} x_{1} x_{3} x_{6}

x_{1} x_{4} x_{5} x_{1} x_{5} x_{6}

x_{1} x_{5} x_{6} = 4 + 3 + 2 + 1 = 10

s i m i l a r l y r_{1} t a k i n g x_{2} w i l l g i v e (3 + 2 + 1) = 6

s i m i l a r l y r_{1} t a k i n g x_{3} w i l l g i v e (2 + 1) = 3

s i m i l a r l y r_{1} t a k i n g x_{4} w i l l g i v e (1) = 1

t o t a l = 6 + 15 + 15 + 10 + 6 + 3 + 1 = 56

Commented by prakash jain last updated on 09/Dec/15

Answer in factorials

csse 1 :^{6} P_{1} (3 r together) = 6

case 2 :^{6} P_{2} (r 2 r 2 s l o t s t o b e f i l l e d i n 6) = 30

case 3 : \frac{P_{3}}{3!} (r r r 3 s l o t s t o b e f i l l e d . 3 d u p l i c a t e s) = 20

Commented by prakash jain last updated on 11/Dec/15

You can also solve this problem with

combination formula .

8 steps need to be taken

SSSSSSSS (5 D, 3 R)

The problem is choosing 3 steps among

8 where R will be placed remaining

position will have D .

total number of ways =^{8} C_{3} = 56

Answered by Rasheed Soomro last updated on 09/Dec/15

T o t a l p o s i t i o n s 24 . L e t e v e r y p o s i t i o n

b e s p e c i f i e d b y c o o r d i n a t e s a s b e l o w :

00 10 20 30

01 11 21 31

02 12 22 32

03 13 23 33

04 14 24 34

05 15 25 35

F r o m 00 3 + 5 = 8 m o v e s a r e p o s s i b l e

F r o m m n

Commented by prakash jain last updated on 09/Dec/15

Consider this

3 r + 5 d

1 d + 3 r + 4 d

2 d + 3 r + 3 d

3 d + 3 r + 2 d

4 d + 3 r + d

4 d + r

2 r + 5 d + r

2 r + 4 d + r + d

2 r + 3 d + r + 2 d

2 r + 2 d + r + 3 d

2 r + d + r + 4 d

and so on .

Commented by Rasheed Soomro last updated on 09/Dec/15

T h^{α} n K^{S} .