Think logically not mathematically [part-10]

Probability Level 4

How many ordered pairs of non-negative integers $(x, y)$ exist for the equation $^{20}C_{y}=^{20}C_{x}$ $note:here^nC_r means \frac{n!}{(n-r)!\times r!}$

11 21 41 10 40 30 20 31

1 solution

Sudoku Subbu
Feb 8, 2015

$if ^{20}C_y=^{20}C_x$ then x+y=20 so the values in ordered pairs are $(0,20)$ $(1,19)$ $(2,18)$ $(3,17)$ $(4,16)$ $(5,15)$ $(6,14)$ $(7,13)$ $(8,12)$ $(9,11)$ and $(10,10)$ beyond this pairs all come under the above pairs as wise we can have the vice versa of the above pairs and also pairs in which x=y

Why can't we have (1,1)? There should be 31 solutions.

Calvin Lin Staff - 6 years, 4 months ago

Yes exactly !

Venkata Karthik Bandaru - 6 years, 3 months ago

Karthik, for this problem we get 41 pairs. They are (0,20)(20,0) (1,19)(19,1) (2,18)(18,2) (3,17)(17,3) (4,16)(16,4) (5,15)(15,5) (6,14)(14,6) (7,13)(13,7) (8,12)(12,8) (9,11)(11,9) (0,0)(1,1) (2,2)(3,3) (4,4)(5,5) (6,6)(7,7) (8,8)(9,9) (10,10) (11,11)(12,12) (13,13)(14,14) (15,15)(16,16) (17,17)(18,18) (19,19)(20,20)

Ritvik Vantipalli - 6 years, 3 months ago

@Ritvik Vantipalli – Oh yes, I get it now. So silly mistake :( I couldn't solve the problem. I marked 31 !

Venkata Karthik Bandaru - 6 years, 3 months ago

I have updated the answer to 31. Please update your solution accordingly.

Calvin Lin Staff - 6 years, 4 months ago

I get 41 pairs. There are 21 pairs of the form $(x, 20 - x)$ , 21 pairs of the form $(x,x)$ , then subtract 1 for double-counting (10,10).

Jon Haussmann - 6 years, 4 months ago

Yes, that is correct. I was thinking that the pairs were unordered because I looked at the above solution, and only added the other $(i,i)$ pairs.

I have updated the answer to 41. Thanks!

Calvin Lin Staff - 6 years, 4 months ago

Think logically not mathematically [part-10]

1 solution

0 pending reports