Menu Close

If-only-downward-motion-along-lines-is-allowed-what-is-the-total-number-of-paths-from-point-P-to-point-Q-in-the-figure-below-




Question Number 17938 by Tinkutara last updated on 12/Jul/17
If only downward motion along lines is  allowed, what is the total number of  paths from point P to point Q in the  figure below?
$$\mathrm{If}\:\mathrm{only}\:\mathrm{downward}\:\mathrm{motion}\:\mathrm{along}\:\mathrm{lines}\:\mathrm{is} \\ $$$$\mathrm{allowed},\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{total}\:\mathrm{number}\:\mathrm{of} \\ $$$$\mathrm{paths}\:\mathrm{from}\:\mathrm{point}\:\mathrm{P}\:\mathrm{to}\:\mathrm{point}\:\mathrm{Q}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{figure}\:\mathrm{below}? \\ $$
Commented by Tinkutara last updated on 12/Jul/17
Commented by ajfour last updated on 13/Jul/17
3 leftdown and 3 rightdown steps  to be chosen.  Number of ways =coeff. of x^3 y^3  in  the expansion of (x+y)^6 =^6 C_3 =20.
$$\mathrm{3}\:\mathrm{leftdown}\:\mathrm{and}\:\mathrm{3}\:\mathrm{rightdown}\:\mathrm{steps} \\ $$$$\mathrm{to}\:\mathrm{be}\:\mathrm{chosen}. \\ $$$$\mathrm{Number}\:\mathrm{of}\:\mathrm{ways}\:=\mathrm{coeff}.\:\mathrm{of}\:\mathrm{x}^{\mathrm{3}} \mathrm{y}^{\mathrm{3}} \:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{expansion}\:\mathrm{of}\:\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{6}} =^{\mathrm{6}} \mathrm{C}_{\mathrm{3}} =\mathrm{20}. \\ $$
Answered by alex041103 last updated on 12/Jul/17
20
$$\mathrm{20} \\ $$
Commented by Tinkutara last updated on 12/Jul/17
Explain.
$$\mathrm{Explain}. \\ $$
Commented by alex041103 last updated on 12/Jul/17
see my answer to Q. 17612
$${see}\:{my}\:{answer}\:{to}\:{Q}.\:\mathrm{17612} \\ $$
Commented by alex041103 last updated on 12/Jul/17
is it ok?
$${is}\:{it}\:{ok}? \\ $$
Commented by alex041103 last updated on 12/Jul/17
Commented by Tinkutara last updated on 13/Jul/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$
Commented by alex041103 last updated on 13/Jul/17
You′re welcome!
$${You}'{re}\:{welcome}! \\ $$
Answered by wullyfalcon last updated on 12/Jul/17
1 continuous move
$$\mathrm{1}\:{continuous}\:{move} \\ $$
Commented by alex041103 last updated on 13/Jul/17
?
$$? \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *