PnC

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

ans is b

b

c

b

a,c

d

just plug in values and check

d

idk about YOU but 15th nov 2025 akshat jain missed an entire case

1. ABC and rest free = 3!

2. ABD and rest free = 3!

3. AC and then BD as a unit with E F are free =3! (as there is only a single valid permutation in BD)

c

c

it moves two steps forward and 1 up

a

c

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

c

d

Inside maane INSIDE

along axis points dont count

c

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

c

a

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

d

a