Question

Please help :
Let X={1,2,3,....10}.Find the number of pairs {A,B} such that A and B are a subsets of X .But A not equal to B .
A intersection B={5,6,8}.

Answer 1

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Answer 2

What do we know about set A?

Answer 3

A and B are elements of X

AnB share the elements 5,6,8

ok, i see it better now

Answer 4

A={5,6,8}
B={1,2,3,4,5,6,7,8,9,10} seems to be a set we can use;

i wonder does the arrangement of elements of X between A and B make a difference?

A={1,2,3,4,5,6,8}
B={5,6,7,8,9,10} would be another suitable set

AxB would seem to be a way to count all the possible ordered pairs to me

5 * * * * x * * * * *
6 * * * * * x * * * *
8 * * * * * * * x * *
1 2 3 4 5 6 7 8 9 10

3*10 - 3 = 27 for the first setup; now to see if the second setup makes a difference



10
9
8 x
7
6 x
5 x
1 2 3 4 5 6 8

it seems like there are always going to be 3 points of the set that are against the rules

6*7-3 = 39 ordered pairs

so i spose arrangement of sets does make a difference

Answer 5

how about A x B

Answer 6

i believe the sets that create the most points is the 7x6

3*10 = 10 , 27
4*9 = 36 , 33
5*8 = 40 , 37
6*7 = 42 , 39

the number of ordered pairs that match the criteria are then

Answer 7

3*10 = 30 :)

Answer 8

we have two generic possibilities
A = * * * * 5 6 * 8 * *
B = * * * * 5 6 * 8 * * , where you can fill in numbers where * is

Answer 9

might as well keep them in order, so the cases are , if you put a number in for * in A, then you must leave it out for B, and vice a versa

Answer 10

, so lets say , for instance, put in 1 . we have two cases (A,B)
A = 1 * * * * 5 6 * 8 * * A = * * * * 5 6 * 8 * *
B = * * * * 5 6 * 8 * * B = 1 * * * 5 6 * 8 * *

Answer 11

, so now you can proceed by singletons , 1,2,3.. then pairs, then triples

Answer 12

RMO 2012 hmm,,did you give? i have seen the question paper..

Answer 13

actually theres more cases thatn i anticipated

Answer 14

are we to find the total number of ordered pairs, as in all the possible sets that can be constructed? or is it the most ordered pairs from a maximal set?

Answer 15

ok thats the best way, count by |A|= 3, |B|=3 , then |A| = 4, |B|=3 , etc

Answer 16

, |A| means cardinality of A

Answer 17

for |A|= 3, |B|=3, there is just one case for (A,B)

Answer 18

now do |A| = 3, |B| = 4, how many (A,B) pairs are there

Answer 19

there are 7 pairs when |A| = 3 , | B| = 4 ,

Answer 20

A and B will surely have 5,6,8
we are left with 1,2,3,4,7,9,10

1st case..

0 elements in A or B
for 0 elements in A,
B can have 1,2,3..or 7 elements..
=>(7C1 + 7C2 ... 7C7)
we multiply this by 2 since A and B are symmetrical i.e. same thing will take place for 0 elements in B
so for 1st case : (7C1 +7C2... 7C7)*2

case 2//
1 element in A or B
=>(6C1 + 6C2..6C6)* 7 *2

case 3..

2 elements in A or B
7C2 * (5C2 + 5C3 .. 5C5) * 2

case 4:
3 elements in A or B
7C3 * (4C3 + 4C4) *2

case 5:

wont be there..

i havent elaborated my solution since i have an intuition it may be wrong..
my ans comes out to be sum of all cases..anyone please check mine..

Answer 21

ok here is an exhaustive list, a schemata

Answer 22

Let |A| = m |B| = k
m= 3 k = 3
m=3 k = 4
m=3 k = 5
...
m=3 k = 7

m = 4 k = 3
m= 4 k = 4
m=4 k = 5
.
.
.
m = 4 k = 7


until you get to m= 7 k = 3... m = 7 k = 7

Answer 23

hey do you hvae the final answer to check ?

Answer 24

I got
7C0 ( 7C0 + 7C1 + 7C2 + 7C3 + 7C4) + 7C1 ( 7C0 + 7C1 + 7C2 + 7C3 + 7C4)
7C2 ( 7C0 + 7C1 + 7C2 + 7C3 + 7C4) + 7C3 ( 7C0 + 7C1 + 7C2 + 7C3 + 7C4)
+ 7C4 ( 7C0 + 7C1 + 7C2 + 7C3 + 7C4)
= ( 7C0 + 7C1 + 7C2 + 7C3 + 7C4) ^2

Answer 25

9801

Answer 26

@eddie2b SRRY i dont have final answer to check

Answer 27

ok

Answer 28

do you understand my explanation ?

Answer 29

@amistre64 , @eddie2b landed on the answer !!
its 2186!!

Answer 30

i made a small editing in the question check:

Answer 31

A is not equal to B , ok

Answer 32

so the answer is 2186, are you sure?

Answer 33

there are \[\large 2^{10}=1024\]distinct subsets of X={1,2,...,10}

Answer 34

yes!!

Answer 35

if we form pairs, then it is just a combination problem with n = 1024 and r = 2

Answer 36

but i dont understand hw to get it !!

Answer 37

are you sure its 2186

Answer 38

\[\large \left(\begin{matrix}1024 \\ 2\end{matrix}\right) =\frac{ 1024 \times 1023 }{ 2 \times 1 }=512 \times 1023\]

Answer 39

yes i even got the solution but i dont understand it!!check question 4

Answer 40

one moment

Answer 41

thanks to @shubhamsrg i searched online for rmo 2012 and gt this bt the solution is nt clear

Answer 42

use slots

Answer 43

A = _ _ _ _ 5 _ 7 8 _ _
B = _ _ _ _ 5 _ 7 8 _ _

Answer 44

?

Answer 45

i understand to solution. why bother about 5, 7 and 8 when they are in BOTH sets? just focus on whether to put the rest of the digits or not.

Answer 46

@sirm3d can u xplain me nt getting it

Answer 47

looks like a typo is in your solutio

Answer 48

consider the digit 1. will you put it in A or in B or not? that gives you three choices.

Answer 49

rest of the elements can be assigned in *3* ways

Answer 50

who wrote this solution?

Answer 51

experts from resonance institute india

Answer 52

aravind, ok start by picking an element , say 1

Answer 53

each of the seven numbers gives you 3 choices, or a total of 3^7=2187 choices.

Answer 54

why cant 1 be in both A and B

Answer 55

A = _ _ _ _ 5 _ 7 8 _ _
B = _ _ _ _ 5 _ 7 8 _ _
for number 1, there are three choices, it is either in A , or B , or neither

Answer 56

what is the ans ? i am getting 2130 ..

Answer 57

@Zarkon the intersection would then include the digit 1 if 1 is in both sets A and B.

Answer 58

@shubham check the pdf i posted its 2186

Answer 59

oh i see the intersection is {5,6,8}...guess I should read

Answer 60

aravind, so start with A = {5,7,8} , B = {5,7,8} , now the number 1 has 3 options , it can go in set A, or it can go in set B , or it can go in neither .

Answer 61

A NOT EQUAL TO B?

Answer 62

so we are looking at the set { 1,2,3,4,6,9,10} ,

Answer 63

wasnt that a ondition?

Answer 64

there is only case where they will be equal , and so you subtract 1 from the final

Answer 65

so the number 1 has three options where it can go , and then for each of that case the next case is number 2 has three options where it can go, and so on , multiply 3^7 , then subtract 1 for the case when you have none in both

Answer 66

how about this, let's label the points in the set