Permutations Practice Test 1

Question : 20 of 40

Marks: +1, -0

There are 10 persons named A, B, ...K. We have the capacity to accommodate only 5. In how many wayscan we arrange them in a line if A is must and G and H must not be included in the team of 5?

$^8P_5$
$^7P_5$
$^7C_3 . (4!)$
$^7C_3 (5!)$

Validate

Solution:

Out of 10 persons, A is in and G and H are out of the team, so we have to select 4 more from 7 remaining. This can be done in

^7C_4

ways. These 5 persons can be arranged in a line in 5! ways.
Hence, the number of possible arrangements is

^7C_4 . 5! = ^7C_3 .(5!)

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