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Question

(D) The total number of one-one functions from a set of $4$ elements to a set of $5$ elements is given by $P(5, 4) = 5 	imes 4 	imes 3 	imes 2 = 12

Vedclass · Accepted Answer

$\frac{1}{20}$

Vedclass · Answer

(D) The total number of one-one functions from a set of $4$ elements to a set of $5$ elements is given by $P(5, 4) = 5 	imes 4 	imes 3 	imes 2 = 120$.We need to find the number of functions satisfying $f(a) + 2f(b) = f(c) + f(d)$,where $f(a), f(b), f(c), f(d)$ are distinct elements from $\{1, 2, 3, 4, 5\}$.Let the values be $x_1, x_2, x_3, x_4$ corresponding to $f(a), f(b), f(c), f(d)$. We need $x_1 + 2x_2 = x_3 + x_4$.Possible sets of values $\{x_1, x_2, x_3, x_4\}$ satisfying the equation:$f(a)$$f(b)$$f(c)$$f(d)$$5$$1$$2$$5$ (Invalid,not one-one)$5$$1$$3$$4$ ($5+2=3+4$,valid)$4$$1$$2$$4$ (Invalid)$4$$2$$3$$5$ ($4+4=3+5$,valid)$1$$3$$2$$5$ ($1+6=2+5$,valid)$2$$3$$4$$1$ ($2+6=4+1$,invalid)For each valid set of values,there are $2$ ways to assign $f(c)$ and $f(d)$ (since $f(c)+f(d)$ is symmetric).Valid cases: $(f(a), f(b), f(c), f(d))$ are $(5, 1, 3, 4), (5, 1, 4, 3), (4, 2, 3, 5), (4, 2, 5, 3), (1, 3, 2, 5), (1, 3, 5, 2)$.Total favorable outcomes $n(A) = 6$.Probability $P(A) = \frac{6}{120} = \frac{1}{20}$.

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

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