In a relay race there are five teams A, B, C, | vedclass

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Question

(A) The total number of ways to choose and arrange the first three teams out of five is given by the permutation formula $^{5}P_{3}$.$^{5}P_{3} = \fra

Vedclass · Accepted Answer

$\frac{1}{10}$

Vedclass · Answer

(A) The total number of ways to choose and arrange the first three teams out of five is given by the permutation formula $^{5}P_{3}$.$^{5}P_{3} = \frac{5!}{(5-3)!} = 5 	imes 4 	imes 3 = 60$.Each of these $60$ outcomes is equally likely,so the probability of each is $\frac{1}{60}$.The event that $A, B,$ and $C$ are the first three to finish means that these three teams occupy the first three positions in any order.The number of ways to arrange $A, B,$ and $C$ in the first three positions is $3! = 3 	imes 2 	imes 1 = 6$.Therefore,the probability is $\frac{3!}{^{5}P_{3}} = \frac{6}{60} = \frac{1}{10}$.

$Column I$	$Column II$
$(A)$ The number of permutations containing the word $ENDEA$ is	$(p)$ $5!$
$(B)$ The number of permutations in which the letter $E$ occurs in the first and the last positions is	$(q)$ $2 \times 5!$
$(C)$ The number of permutations in which none of the letters $D, L, N$ occurs in the last five positions is	$(r)$ $7 \times 5!$
$(D)$ The number of permutations in which the letters $A, E, O$ occur only in odd positions is	$(s)$ $21 \times 5!$

In a relay race there are five teams $A, B, C, D$ and $E$. What is the probability that $A, B$ and $C$ are the first three to finish (in any order)? (Assume that all finishing orders are equally likely.)

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