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Let $X$ and $Y$ denote the sets containing 2 and 20 distinct objects respectively and $F$ denote the set of all possible functions defined from $X$ to $Y$. Let $f$ be randomly chosen from $F$. The probability of $f$ being one-to-one is ______.
in Set Theory & Algebra 3.3k views
3

Let

#elements in co-domain = m

#elements in domain = n

#one-to-one functions = P( m, n )

#total functions = mn

probability =  P(m,n ) / mn

4 Answers

42 votes
 
Best answer
For a function, the first element in $X$ has $20$ choices (to map to) and the second element also has $20$ choices. For a one-to-one function the second element has only $19$ choices left after $1$ being taken by the first. So, required probability

$=\frac {(20 \times 19)} {(20 \times 20)} = 0.95$

edited by
2

This might help ... 

(No of one to one functions )$/$ ( No of Total possible functions ) = $(20_{p_{2}} /20^{2})$ = $0.95$

5 votes

Total functions from X to Y = [Order(Y) ]order(x)

and number of one-one functions = 20 P 2

so probability = number of one one functions / total number of functions = 20*19/20*20 = 0.95


edited by
2 votes

∣X∣ = 2 , ∣Y∣ = 20

F: X ⩶>Y

Total number of possible functions from X to Y =  ∣Y∣^( ∣X∣) = 20^2 =400

Total number of possible one to one functions from X to Y = 20c1 ⨉ 19c1 =380

Probability(one-to-one) =380 / 400 = 0.95

The correct answer is 0.95.

2
Total functions from m element set to n element set is $n^{m}$ and number of one-one functions are nPm.
0 votes

Answer : 0.95

for function X to Y, where |X| = m & |Y|=n

total one-to-one functions = P(n,m)

total  functions = n^m .

probability(one-to-one) = total  one-to-one functions / total  functions

probability(one-to-one)  = P(20,2) / 20^2 =  (19*20) / 400 = 0.95

Answer:

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