Out of 10 persons sitting at a round table. Three persons are selected at random one after the other. The chance that no two of the selected are together.

\frac{1}{2}

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\frac{2}{5}

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\frac{5}{9}

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\frac{5}{12}

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Solution

The correct option is D $\frac{5}{12}$
Out of 10 persons 3 can be selected in $10_{P_{3}}$ ways = 10.9.8 = 720. First person in 10 ways. Other 2 in $7_{C_{2}}$ ways in which 2 are together in 6 ways. Hence $P (E) = \frac{10 \times (7_{C_{2}} - 6) \times 2}{720} = \frac{5}{12}$

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Out of 10 persons sitting at a round table. Three persons are selected at random one after the other. The chance that no two of the selected are together.

The correct option is D 512Out of 10 persons 3 can be selected in 10P3 ways = 10.9.8 = 720. First person in 10 ways. Other 2 in 7C2 ways in which 2 are together in 6 ways. Hence P(E)=10×(7C2−6)×2720=512

The correct option is D $\frac{5}{12}$
Out of 10 persons 3 can be selected in $10_{P_{3}}$ ways = 10.9.8 = 720. First person in 10 ways. Other 2 in $7_{C_{2}}$ ways in which 2 are together in 6 ways. Hence $P (E) = \frac{10 \times (7_{C_{2}} - 6) \times 2}{720} = \frac{5}{12}$