The probability that a randomly chosen one-one function from the set {a, b, c, d} to the set {1, 2, 3, 4, 5} satisfies f(a) + 2f(b) $-$ f(c) = f(d) is :
Solution
Number of one-one function from $\{a, b, c, d\}$ to set $\{1,2,3,4,5\}$ is ${ }^{5} P_{4}=120 n(s)$.
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The required possible set of value (f(a), $f(b), f(c), f(d))$ such that $f(a)+2 f(b)-f(c)=f(d)$ are $(5,3,2,1),(5,1,2,3),(4,1,3,5),(3,1,4,5)$, $(5,4,3,2)$ and $(3,4,5,2)$
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$\therefore n(E)=6$
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$\therefore $ Required probability $=\frac{n(E)}{n(S)}=\frac{6}{120}=\frac{1}{20}$
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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