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expeople1 [14]
2 years ago
10

Permutation:

Mathematics
2 answers:
Minchanka [31]2 years ago
4 0

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=\dfrac{n!}{r!(n-r)!}

  • Put values

\\ \rm\hookrightarrow ^7C_5

\\ \rm\hookrightarrow \dfrac{7!}{5!(7-5)!}

\\ \rm\hookrightarrow \dfrac{7!}{5!2!}

\\ \rm\hookrightarrow \dfrac{7\times 6\times 5!}{5!(2)}

\\ \rm\hookrightarrow \dfrac{7(6)}{2}

\\ \rm\hookrightarrow \dfrac{42}{2}

\\ \rm\hookrightarrow 21ways

Kruka [31]2 years ago
3 0

Formula for permutation is:

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

<h3><u>Calculation</u>, </h3>

Plug in the values:

Required number of ways = ²C7

  • n = 7
  • r = 5

\sf \frac{7!}{5!(7 - 5)!}

\sf \frac{7 \times 6 \times  \cancel{5!}}{ \cancel{5! }\times 2!}

\sf \frac{7 \times 6 }{ 2!}

\sf \frac{42 }{ 2!}

\sf \frac{42 }{ 2}  = 21

<em>Thus, </em><em>The</em><em> </em><em>players</em><em> </em><em>can</em><em> </em><em>be</em><em> </em><em>selected</em><em> </em><em>in</em><em> </em><em>21</em><em> </em><em>different</em><em> </em><em>ways</em><em>!</em><em>!</em><em>~</em>

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The answers will be:

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