CMI2014-A-01

Question

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.

• A. $260$
• B. $340$
• C. $720$
• D. $440$

Comments

Basic rule of PnC when we do multiplication then all possible combination are automatically counted (Cartesian product).

Take a small example

Que: From 3 batsman and 2 bowlers a team of 3 has to be selected such that there is at least one bowler.

If you do lets select one bowler and then remaining 2 from other 4 : $\binom{2}{1} * \binom{4}{2}$

If you look carefully it is like forming a set of size 3 and satisfying the criteria. {${\binom{2}{1}(first element),\binom{4}{2}}$(second and third element)}.  Here if in remaining 4C1 there is a bowler, then it will form all possible combination (first B1 is selected and then B2 or first B2 is selected and then B1) but it is same. This is the factor of over counting.

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Always convert "at least"  to "exactly" and then think about the problem. It will give you better insight.

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awesome.. thanks a lot.

Answer 1

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

Comments

sir Why not 5C2*4C2*2C1*6C1 = 720

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why can't we solve like   (5 C 2 ∗ 4 C 2 ∗ 2 C 1 (3 C 1 + 2 C 1 + 1 C 1) ) ?? First selecting minimum number of players and then selecting 1 member from remaining type...

Answer 2

Answer A) 260

Comments

your method?

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The best answer is the good method.

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@Srestha ma'am,

What is wrong in this approach, selecting first necessary players, 5C2*4C2*2C1 =120 ways, then after this selection 6 players are left so selecting 1 among them 6C1, so 6*120=720?.

I understand how 260 would be right but what is going wrong in this approach?

Thank you.

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I guess because selecting 6 out of 11 without any restriction is 11C6 which is less than 720.

Answer 3

Answer 4

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