Permutations Practice Test 1

Question : 30 of 40

Marks: +1, -0

Ten persons are to speak at a meeting. The number of ways in which it can be arranged if A wantsto speak before B and B wants to speak before C, is:

$10!/3$
7!
$10!/6$
7! × 3!

Validate

Solution:

Number of arrangements of 10 persons is (10)!. In these arrangement A,B,C are also permuted 3!= 6 times, but ABC has a fixed order. Hence, the number of required arrangements is

(10)!/6

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