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find in how many ways can two admirals, three generals and 4 lieutenants can be seated in a row of nine seats if the admirals must sit at each end?
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Sagot :

Answer:

They can be seated in 10,080 ways.

Step-by-step explanation:

Arrangements of n elements:

The number of possible arrangements of n elements is given by:

[tex]A_n = n![/tex]

In this question:

In each end, the two admirals, so 2 possible outcomes.

In the middle seats, arrangements of 7 elements(3 generals, 4 lieutenants). So

[tex]2A_{7} = 2*7! = 10080[/tex]

They can be seated in 10,080 ways.

Abidemiokinidn

The number of ways they can be seated is 10,080 ways.

Permutations and combination

The number of possible ways fof arranging n elements is given as:

P = n!

According to the question, there are 2 admirals! 3 generals and 4 lieutenant.

In each end, the two admirals, so 2 possible outcomes.

In the middle seats, arrangements of 7 elements(3 generals, 4 lieutenants). Hence the required number of combination will be given as:

Number of ways = 2*7!

Number of ways = = 100802

Hence the number of ways they can be seated is 10,080 ways.

Learn more on permutations here: https://brainly.com/question/1216161

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