Ten persons, amongst whom are $A, B$ and $C$ are to speak at a function. The number of ways in which it can be done if $A$ wants to speak before $B$ , and $B$ wants to speak before $C$ is

\frac{10!}{6}

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8!

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\frac{8!}{3}

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3!

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Solution

The correct option is D $\frac{10!}{6}$
Places for A, B and C can be chosen in $^{10} C_{3}$ ways
Remaining 7 can be selected in $7!$ ways
So total $=^{10} C_{3} \times 7!$
$= \frac{10! .7!}{3! .7!}$
$= \frac{10!}{6}$
Hence, option 'A' is correct.

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Ten persons, amongst whom are A, B and C to speak at a function. The number of ways in which it can be done if A wants to speak before B and B wants to speak before C is

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Q. 15 persons which include A, B, C are to speak at a function. Find the number of ways in which it can be done, if A wants to speak before B and B wants to speak before C.

A, B, C

are three persons among

7

persons who speak at a function. The number of ways in which it can be done. If '

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Ten persons, amongst whom are A,B and C are to speak at a function. The number of ways in which it can be done if A wants to speak before B, and B wants to speak before C is

The correct option is D 10!6Places for A, B and C can be chosen in 10C3 waysRemaining 7 can be selected in 7! waysSo total =10C3×7!=10!.7!3!.7!=10!6Hence, option 'A' is correct.

Ten persons, amongst whom are $A, B$ and $C$ are to speak at a function. The number of ways in which it can be done if $A$ wants to speak before $B$ , and $B$ wants to speak before $C$ is

The correct option is D $\frac{10!}{6}$
Places for A, B and C can be chosen in $^{10} C_{3}$ ways
Remaining 7 can be selected in $7!$ ways
So total $=^{10} C_{3} \times 7!$
$= \frac{10! .7!}{3! .7!}$
$= \frac{10!}{6}$
Hence, option 'A' is correct.