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Let X={a1,a2......a6} and Y={b1,b2,b3}.Find number of functions f from x to y such that it is onto and there are exactly three elements x in X such that  f(x)=b1.

7 years ago


Answers : (1)


In mathematics, a function f is said to be surjective or onto, if its values span its whole codomain; that is, for every y in the codomain, there is at least one x in the domain such that f(x) = y .

We can choose any 3 of a1,a2....,a6, say c1,c2,c3. f(c1)=f(c2)=f(c3)=b1.

[For example, lets say f(a2),f(a5),f(a6) are equal to b1. Then, c1,c2,c3 are a2,a5 and a6]

c1,c2 and c3 can be chosen in 6C3 = 20 ways.

After removing c1,c2 and c3 from a1,a2,....a6, we will have three more a's left. Let these be d1,d2 and d3.

[In the above example, d1,d2 and d3 will be a1,a3, and a4]

We need f(d1),f(d2) and f(d3) to be either b2 or b3, but at least one of them should be b2 and at least one of them should be b3 because the function f is onto.

So the possible ways we can do this = 2*2*2 - 2 = 6

[Because we can assign f(d1) as b2 or b3. There are 2 ways of doing this. Similarly, we can assign f(d2)=b2 or b3 and same for f(d3). There are 2*2*2=8 ways of doing this. Then we subtract the two cases where all three are b1 or b2]

So total number of such functions = 20 * 6 = 120.

Let me know if you face any problem in understanding this explanation, especially if you have not yet done Permutations and combinations.


7 years ago

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