3 should be at third pop, before that we have to do two pop operations.
I) fix the first pop operation as 1, then
i) fix the second pop operation as 2, ===> remaining three elements i.e., 4,5 and 6 lead to $\frac{\binom{2n}{n}}{n+1} = \frac{\binom{6}{3}}{4} = 5 $
ii) fix the second pop operation as 4, ===> remaining three elements i.e., 5,6, and 2
5 and 6 yet to be pushed but remember 2 is in stack ===> lead to $\frac{\binom{2n}{n}}{n+1} = \frac{\binom{6}{3}}{4} = 5 $
iii) fix the second pop operation as 5, ===> if 2$^{nd}$ pop is 5, then third pop should be 4 ===> invalid case
WHY ?
4 is pushed after 3, 5 is pushed after 4 ===> 4 is lie between 3 and 5, then without poping 4, you can't pop 3.
iv) fix the second pop operation as 6, ===> if 2$^{nd}$ pop is 6, then third pop should be 5 ===> invalid case
II) fix the first pop operation as 2, then
i) fix the second pop operation as 1, ===> remaining three elements i.e., 4,5 and 6 lead to $\frac{\binom{2n}{n}}{n+1} = \frac{\binom{6}{3}}{4} = 5 $
ii) fix the second pop operation as 4, ===> remaining three elements i.e., 5,6, and 1
5 and 6 yet to be pushed but remember 1 is in stack ===> lead to $\frac{\binom{2n}{n}}{n+1} = \frac{\binom{6}{3}}{4} = 5 $
iii) fix the second pop operation as 5, ===> if 2$^{nd}$ pop is 5, then third pop should be 4 ===> invalid case
WHY ?
4 is pushed after 3, 5 is pushed after 4 ===> 4 is lie between 3 and 5, then without poping 4, you can't pop 3.
iv) fix the second pop operation as 6, ===> if 2$^{nd}$ pop is 6, then third pop should be 5 ===> invalid case
III) fix the first pop operation as 4, then
i) fix the second pop operation as 5, ===> remaining three elements i.e., 1,2 are in stack and 6 is yet to be pushed
then possible, 1 should be after 2 and 6 can be any where ===> 3 orders
i.e., 2 1 6 , 2 6 1 and 6 2 1
iv) fix the second pop operation as 6, ===> if 2$^{nd}$ pop is 6, then third pop should be 5 ===> invalid case
Note you can't pop either 2 or 1 as second pop, due to 3 is pushed on those 2 elements
IV) fix the first pop operation as 5, then second pop operation should be 4, to get third pop operation is 3
===> remaining three elements i.e., 1,2 are in stack and 6 is yet to be pushed
then possible, 1 should be after 2 and 6 can be any where ===> 3 orders
i.e., 2 1 6 , 2 6 1 and 6 2 1
Total = I + II + III + IV = ( 5+5 = 10 ) + ( 5+5 = 10 ) + 3 + 3 = 26