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  • A team of 6 players is to be selected from a group of 10 players. In how many ways can the team be formed if the order of selection does not matter?<div class='qna-op

A team of 6 players is to be selected from a group of 10 players. In how many ways can the team be formed if the order of selection does not matter?

Option: 1

110


Option: 2

210


Option: 3

320


Option: 4

520


Answers (1)

best_answer

To find the number of ways a team of 6 players can be formed from a group of 10 players, where the order of selection does not matter, we can use the concept of combinations.

The number of combinations can be calculated using the formula for combinations:

\mathrm{ C(n, r)=n ! /(r ! \times(n-r) !) }

Where:
\mathrm{n} is the total number of players in the group ( 10 in this case)

\mathrm{r} is the number of players to be selected for the team (6 in this case)

\mathrm{!} represents the factorial function

Plugging in the values into the formula:

\mathrm{ C(10,6)=10 ! /(6 ! \times(10-6) !) }

Simplifying the factorial expressions:

\mathrm{ 10 !=3,628,800 }

\mathrm{ 6 !=720 }

\mathrm{ (10-6) !=24 }

Substituting these values into the formula:

\mathrm{ C(10,6)=3,628,800 /(720 \times 24)=3,628,800 / 17,280=210 }

Therefore, there are 210 ways to form a team of 6 players from a group of 10 players, where the order of selection does not matter.

Hence option 2 is correct.

Posted by

Shailly goel

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