ask 6,10,12,17,18,22,25,27,28,31,32,33,
ask method 3,4
Alternate numbers are 1
the other numbers are prime numbers
so you can have 2 cases
1p1p1p1p1p
p1p1p1p1p1
ans is b
ans b
for rectangles, its nc2*nc2 (just choose any two lines from each side you get a rectangle)
for squares is 1^2 + 2^2 ….. 8^2
ask method
b
c
a,c
d
just plug in values and check
idk about YOU but 15th nov 2025 akshat jain missed an entire case
ABC and rest free = 3!
ABD and rest free = 3!
AC and then BD as a unit with E F are free =3! (as there is only a single valid permutation in BD)
it moves two steps forward and 1 up
a
111, 112,113…..119 = 9
122,123,…..129 = 8
…
188,189 = 2
199 = 1
from 200s we start from 8 and get to 1
this goes on till we get to 999 which is only 1
so ans = (1+2…9) + (1+2…8) +…. +1
= 165
Inside maane INSIDE
along axis points dont count
30030 = 2x3x5x7x11x13
make three bucket for a,b,c
each number out of 2,3,5,7,11,13 gets a choice to go into any 1 of these boxes
now we wont get a<b<c this way but there will always be some increasing order and thus they can be arranged accordingly, except in the cases where all the numbers get into just 1 bucket
So ans = 3^6 -3
n=4 as you SHOULDVE calculated correctly
empty seat and the two beside them are not available
so 4c2 - 2 (2 cases mei the other 2 dudes would be consecutive)
so n = 4
ans is abcd
note they are sitting at a circular table
there are 12 guests NOT INCLUDING master and mistress
just seat the master and mistress and they act as reference
consider the two people who wanna sit together as a unit
choose which side they'll sit on (2c1)
they have 5 locations to choose from each side (5c1)
they can be permuted within the unit by 2!
the rest can sir wherever so 10!
so total is 2x5x2x10! = a
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