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51 votes
51 votes
The number of onto functions (surjective functions) from set $X = \{1, 2, 3, 4\}$ to set $Y=\{a,b,c\}$ is ______.
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15 Answers

7 votes
7 votes

Surjective function : when co- domain = range so X={1,2,3,4} to Y={a,b,c} i

s from 4 element choose any two and map to element 'a' of Y so 4c2 = 6 posibility

from remaining 2 choose 1 map to any other element to Y = 2c1= 2

remaing one may directly one way     only so 6*2*1= 12 ways

now from 4 element choose any two and map to element 'b' of Y so 4c2 = 6 posibility

from remaining 2 choose 1 map to any other element to Y = 2c1= 2 r

emaing one may directly one way     only so 6*2*1= 12 ways

from 4 element choose any two and map to element 'b' of Y so 4c2 = 6 posibility

from remaining 2 choose 1 map to any other element to Y = 2c1= 2

remaing one may directly one way only so 6*2*1= 12 ways

total ways = 12+12+12=36

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5 votes
5 votes

My approach to solve this is some story, sorry plz!

Let, 'a' our boss so he can rule on 2 elements possibly {(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)} (try to make pair only in this direction 1-->2-->3-->4)

now remaining two slaves 'b' and 'c' can have only one of two elements.

so this way total 12 ways possible for ex { [a-1,2 , b-3 , c-4 , a-1,2 , b-4 , c-3],...}  12 possibilities 

Now, it's time to make 'b' as a boss and 12 possibilities 

Now, it's time to make 'c' as a boss and 12 possibilities 

so, total 12+12+12=36

P.S. this may sound weired but story is the medicine to remember something!

5 votes
5 votes

Another way to solve this problem. 

$x_{1}$ $x_{2}$ $x_{3}$ $x_{4}$

Total (Row wise)

a a b c $\frac{(4!)}{(2!)(1!)(1!)}$
a b b c $\frac{(4!)}{(2!)(1!)(1!)}$
a b c c $\frac{(4!)}{(2!)(1!)(1!)}$
     

Final Total

3*(3*4) = 36
4 votes
4 votes

One way of solving it can be:

Choose two numbers from X to be mapped to one element in Y -> $\binom{4}{2} * 3$
Now we are left with 2 elements in X and 2 elements in Y we can have only two possible ways.

Total =>  $\binom{4}{2} * 3$ * 2 = 36

Answer:

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