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For the inter-hostel six-a-side football tournament, a team of $6$ players is to be chosen from $11$ players consisting of $5$ forwards, $4$ defenders and $2$ goalkeepers. The team must include at least $2$ forwards, at least $2$ defenders and at least $1$ goalkeeper. Find the number of different ways in which the team can be chosen.

  1. $260$
  2. $340$
  3. $720$
  4. $440$
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Best answer
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There are three ways to choose 6 Players.

  1.  $^5C_3*^4C_2*^2C_1=120.$
  2.  $^5C_2*^4C_2*^2C_2=60.$
  3.  $^5C_2*^4C_3*^2C_1=80.$

So total No. of ways is $260.$

Correct Answer: A

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Answer A) 260
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I don’t know why anytime i see at least question in mind always came

Total – {something}

so here also i see at least again i think can we solve in the above way . Then i try and here is the result

11C6 – {5C0*4C4*2C2 + 5C1*4C4*2C1 + 5C1*4C3*2C2 + 4C0*5C5*2C1 + 4C0*5C4*2C2 + 4C1*5C5*2C0 + 4C1*5C4*2C1 + 4C1*5C3*2C2 + 2C0*5C4*4C2 + 2C0*5C3*4C3 + 2C0*5C2*4C4}

Note : I didn’t include 2C0*5C5*4C1 because 6th term is same as this

462 – 202 = 260 answer
Answer:

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