Question

Permutation:

A college has 7 good badminton players.In how many ways can a team of 5 players be selected??​

Answer 1

Here one player cannot be repeated so repeating is not allowed.

• Hence we use combination here

• n=7
• r=5

Total ways

`\\ \rm\hookrightarrow ^nC_r=(n!)/(r!(n-r)!)`

• Put values

`\\ \rm\hookrightarrow ^7C_5`

`\\ \rm\hookrightarrow (7!)/(5!(7-5)!)`

`\\ \rm\hookrightarrow (7!)/(5!2!)`

`\\ \rm\hookrightarrow (7* 6* 5!)/(5!(2))`

`\\ \rm\hookrightarrow (7(6))/(2)`

`\\ \rm\hookrightarrow (42)/(2)`

`\\ \rm\hookrightarrow 21ways`

Answer 2

Formula for permutation is:

`\boxed{ \tt^(n) C_(r) = (n!)/(r!(n - r)!) }`

Calculation,

Plug in the values:

Required number of ways = ²C7

• n = 7
• r = 5

`\sf (7!)/(5!(7 - 5)!)`

`\sf \frac{7 * 6 * \cancel{5!}}{ \cancel{5! }* 2!}`

`\sf (7 * 6 )/( 2!)`

`\sf (42 )/( 2!)`

`\sf (42 )/( 2) = 21`

Thus, The players can be selected in 21 different ways!!~