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Question: In how many ways can a team of \(11\) players be formed out of \(25\) years, If six out of them are ...

In how many ways can a team of 1111 players be formed out of 2525 years, If six out of them are always to be included and five always to be excluded ?
A) 20202020
B) 20022002
C) 20082008
D) 80028002

Explanation

Solution

In this question we have to select the number of players according to the given data in the question. As we can see, the above question is related to Permutation and Combination. The combination is a way of selecting items from a collection or we can say the grouping of outcomes.
So in this question we will use the formula for the number of ways for selecting rr things from nn group of people. The formula of combination is given as;
nCr=n!r!(n−r)!^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}.
So we will use this formula to solve the given problem.
We will first include the six players and then we have to exclude the five players. So Total of these players are
6+5=116 + 5 = 11
We will subtract this from the total players and then we apply the formula.

Complete step by step answer:
In the given question we have total number of players
2525
Out of them we have to include 66 players and exclude 55 players always, total of these are 6+5=116 + 5 = 11 players.
So we have left with number of players i.e.
25−11=1425 - 11 = 14 players.
Now we should note that we have to form a total of 1111 players.
We have already included 66 players in the above, so we have included five more players out of the remaining 1414 players.
So, five players are to be chosen from 1414, we have :
14C5^{14}{C_5}
By comparing from the formula we have here,
n=14,r=5n = 14,r = 5
We can break the value by applying the formula i.e.
14!5!(14−5)!\dfrac{{14!}}{{5!(14 - 5)!}}
We will now break the values and simplify this:
14×13×12×11×10×9!5×4×3×2×9!=14×13×12×11×105×4×3×2\dfrac{{14 \times 13 \times 12 \times 11 \times 10 \times 9!}}{{5 \times 4 \times 3 \times 2 \times 9!}} = \dfrac{{14 \times 13 \times 12 \times 11 \times 10}}{{5 \times 4 \times 3 \times 2}}
On multiplying it gives us the value
14×13×11=200214 \times 13 \times 11 = 2002
Hence, the correct option is (B) 20022002.

Note:
We should note that for selection purposes we use combination and for arranging the values we use the permutation. We should know that if the order does not matter then we use the combination formula as in the above question, but if the order does matter then we use the permutation formula. The value of permutation is denoted by nPr^n{P_r}.
The formula of permutation is:
n!(n−r)!\dfrac{{n!}}{{\left( {n - r} \right)!}}