Question

Three children,each accompanied by a guardian,seek admission in a school.The principal wants to interview all the 6 people one after another subject to the condition that no child is interviewed before the guardian.In how many ways can this be done?

Answer 1

@ganeshie8 @genius12

Answer 2

3! x3! =36 ways

Answer 3

lol? @dlearner where'd you get that from? 0,o

Answer 4

ok...so there are 3 guardians and 3 kids

Answer 5

@dlearner Please do give hints..don't give answers

Answer 6

but princi wants to go through the guardians first

Answer 7

there are 3 guardians..so 3 ways in which the pricni can meet thm....do you get that?

Answer 8

@dlearner You got that part right, but you assumed that the three "children" and three "Guardians" are distinguishable when this might not be the case. If they are distinguishable then there is (3!)^2 ways but if they are not then there is only 1 way.

Answer 9

The 3 "children" and 3 "guardians" may suggest that the children's order or guardians' order doesn't matter to the principal but if the names for each individual were given then it would be a different story.

Answer 10

So if the children/guardians are distinguishable; 36 ways. If not then 1 way. @DLS

Answer 11

oh yeah..you r right..i get it

Answer 12

60
90
120
180
are the options..

Answer 13

oops :D

Answer 14

lol......?

Answer 15

@DLS you sure there is no tricks or anything...

Answer 16

no child is interviewed before the guardian

Answer 17

meaning each guardian before his child

Answer 18

G1,G2,G3 are the 3 guardians and
C1,C2,C3 are the 3 children..
Many cases can arise..
we might considering binding G1 C1 in 1 group etc..and permuting it..ive actually forgotten PNC XD

Answer 19

all possible permutations of interviewing 6 ppl = 6!

Answer 20

In half of them, G1 will be after C1

Answer 21

so, in 6!/2 permutations, G1 will be before C1.

get the idea ?

Answer 22

hmmp

Answer 23

gr8 stuff ;)

Answer 24

out of 6!/2,

in half of above, G2 will be after C1
so, in 6!/4 permutations, G1 will be before C1, and G2 will be before C2

Answer 25

6!/8 is your answer

Answer 26

i had a type in my previous reply :-

out of 6!/2,

in half of above, G2 will be after \(\color{red}{C2}\)
so, in 6!/4 permutations, G1 will be before C1, and G2 will be before C2

Answer 27

seems correct! but why didn't we subtract 8 then?

Answer 28

i mean out of 6! (total permutations) - 3 cases(C1G1,C2G2,C3G3) ? I mean this

Answer 29

answer is 120 BTW not 90..

Answer 30

6!/8 is 93. something not in options ??

Answer 31

omg i forgot that the question is talking about the RESPECTIVE guardian lol................

Answer 32

it is 90 @dlearner

Answer 33

the way i was thinking was that no guardian could go before a child lol -.-

Answer 34

haha

Answer 35

i mean no child can go before a guardian* lol

Answer 36

yeah..a little mistake @DLS

Answer 37

we can subtract the 3 cases aswell.. but it wud be painful. division is simple here

Answer 38

we still didn't get right answer..and can anybody tell what is wrong with the statement I made above?

Answer 39

720-3= 717 ? :|

Answer 40

is that the way u preparing for JEE ha ?

Answer 41

:O

Answer 42

look at the attached file - brute force of 720 permutations and pruning the permutations in which children are before their respective guardians. 90 is your answer.. 120 is incorrect.

Answer 43

why is 720-3 incorrect?

Answer 44

from where u got 3 ?

Answer 45

3 cases..
C1G1
C2G2
C3G3 are the cases to be discarded..right? :/

Answer 46

okay i guess i got it :P

Answer 47

3 cases :-
1) first case when C1 is before G1
this gives u many permuations. work them

Answer 48

cool :) when u free try the subtraction method. should give u 90