How many ways a team of 11 players can be formed out of 25 players if 6 out of them are always to be included and 5 are always to be excluded?

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Answer

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Hint: Selection of r different things from n different things is given by the expression $c_{r}^{n}=\dfrac{n!}{r!\left( n-r \right)!}$. For including one person, include him in the team and select other players from the remaining players. For excluding, remove one person from total players present and select total players required from the remaining players.Complete step-by-step answer:As we know the number of ways of selecting ‘r’ things from ‘n’ different objects can be given by relation:$c_{r}^{n}=\dfrac{n!}{r!\left( n-r \right)!}$ ………………… (i)So, hence we need to select a football time of 11 players from 15 players. Hence using the number of teams formed with 11 players as $c_{11}^{15}$ .Now, using the equation (i), we can simplify it as$c_{11}^{15}=\dfrac{15!}{4!11!}$ As, we can write 15! as $15\times 14\times 13\times 12\times 11!$ So, we get$\begin{align} & c_{11}^{15}=\dfrac{15\times 14\times 13\times 12\times 11!}{4!11!} \\ & =\dfrac{15\times 14\times 13\times 12}{4\times 3\times 2\times 1}=105\times 13 \\ \end{align}$ = 1365Hence, there are 1365 ways to form a football team with 11 players from 15 players.i) Include one particular player:As it is already given that the football team consists of 11 players. Now, if we want to include one particular player out of 15, that player will always be in 11 selected players. It means that we have selected one particular player, out of the remaining 14 players. So, we can select 10 players out of 14 players with the help of equation (i) as $c_{10}^{14}$. So, we get $\begin{align} & c_{10}^{14}=\dfrac{14!}{10!4!}=\dfrac{14\times 13\times 12\times 11\times 10!}{10!\times 4\times 3\times 2\times 1} \\ & =91\times 11 \\ \end{align}$ = 1001Hence, there will be 1001 ways to select 11 players from 15 plays if we want to include one particular player always.ii) Exclude one particular player:As, we need to select 11 players for the football time from given 15 players but here we are not considering one particular player for selection as we need to exclude him/her. It means we need to select 11 players from the remaining 14 players.Hence, it can be given with the help of a relation given in the equation as $c_{11}^{14}$. So, we can solve it as:$c_{11}^{14}=\dfrac{14!}{3!11!}=\dfrac{14\times 13\times 12\times 11!}{3\times 2\times 1\times 11!}=14\times 26$ = 364Hence, we can form a football team of 11 players from 15 players by 365 ways, if we are excluding one particular player.Note: One may go wrong with the cases of excluding and including. Just do the process of exclusion or inclusion at the very starting of the solution, either select 10 players from 15 players or select 11 players from 15 players then try to add or subtract some cases, which will be a very complex approach and he/she may lose some of the cases. So, be careful with these kinds of questions and do exclusion or inclusion at the very starting.Don’t use the $P_{r}^{n}$ formula for selecting the players. As it will select the players and arrange them as well. So, we need to do only the selection part, not arrangement of the players. Hence, don’t arrange them using another approach as well. Be clear with the concepts of selection and arrangement.

Answer

Verified

Hint: 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 $r$ things from $n$ group of people. The formula of combination is given as;$^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$. 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 $6 + 5 = 11$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$25$Out of them we have to include $6$ players and exclude $5$ players always, total of these are $6 + 5 = 11$ players.So we have left with number of players i.e.$25 - 11 = 14$ players.Now we should note that we have to form a total of $11$ players.We have already included $6$ players in the above, so we have included five more players out of the remaining $14$ players.So, five players are to be chosen from $14$, we have :$^{14}{C_5}$By comparing from the formula we have here, $n = 14,r = 5$We can break the value by applying the formula i.e.$\dfrac{{14!}}{{5!(14 - 5)!}}$We will now break the values and simplify this:$\dfrac{{14 \times 13 \times 12 \times 11 \times 10 \times 9!}}{{5 \times 4 \times 3 \times 2 \times 9!}} = \dfrac{{14 \times 13 \times 12 \times 11 \times 10}}{{5 \times 4 \times 3 \times 2}}$On multiplying it gives us the value $14 \times 13 \times 11 = 2002$

Hence, the correct option is (B) $2002$.

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 $^n{P_r}$.The formula of permutation is:$\dfrac{{n!}}{{\left( {n - r} \right)!}}$

Correct Answer: B

Solution :
[b] Since 5 players are always to be excluded and 6 players always to be included, therefore 5 players are to be chosen form 14. Hence. Required number of ways is \[^{14}{{C}_{5}}\]=2002.

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Correct Answer: C

Solution :

[c] The number of ways the candidate can choose questions under the given conditions is enumerated below.

Group 1	Group 2	Number of ways
4	2	\[{{(}^{5}}{{C}_{4}}){{(}^{5}}{{C}_{2}})=50\]
3	3	\[{{(}^{5}}{{C}_{3}}){{(}^{5}}{{C}_{3}})=100\]
2	4	\[{{(}^{5}}{{C}_{2}}){{(}^{5}}{{C}_{4}})=50\]
	Total number of ways	200

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Correct Answer: A

Solution :
[a] Matches whose prediction are correct can be selected in \[^{20}{{C}_{10}}\]ways. Now each wrong prediction can be made in 2 ways. Thus, the total number of ways is \[^{20}{{C}_{10}}\times {{2}^{10}}\]

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Correct Answer: C

Solution :
[c] Out of 10 points let n points be collinear. Then the number of triangles is \[^{10}{{C}_{3}}{{-}^{n}}{{C}_{3}}=110\] or \[\frac{10\times 9\times 8}{6}-\frac{n(n-1)(n-2)}{6}=110\] or \[n(n-1)(n-2)=60\] or \[n=5\]

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Correct Answer: C

Solution :

[c] (i) Miss C is taken

[a] B included \[\Rightarrow \]A excluded \[{{\Rightarrow }^{4}}{{C}_{1}}{{\times }^{4}}{{C}_{2}}=24\]

[b] B excluded \[{{\Rightarrow }^{4}}{{C}_{1}}{{\times }^{5}}{{C}_{3}}=40\]

Miss C is not taken

\[\Rightarrow \]B does not come\[{{\Rightarrow }^{4}}{{C}_{2}}{{\times }^{5}}{{C}_{3}}=60\]

\[\Rightarrow \]Total=124

Alternate method:

Case I:

Mr. B is present

\[\Rightarrow \]A is excluded and C included

Hence, the number of ways is \[^{4}{{C}_{2}}^{4}{{C}_{1}}=24\].

Case II:

Mr. 'B' is absent

\[\Rightarrow \]No constraint

Hence, the number of ways is \[^{5}{{C}_{3}}^{5}{{C}_{2}}\]=100.

\[\therefore \]Total = 124.

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Correct Answer: B

Solution :
[b] Dice is marked with numbers 1, 2, 3, 4, 5, 6. If the sum of dice in three throws is 11, then observation must be 1, 4, 6;...1, 5, 5;...2, 3, 6;...2, 4, 5;...3, 4, 4. We can get this observation in 3!+3!/2!+3!+3!/2!+3!/2!=27 ways.

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Correct Answer: A

Solution :
[a] Obviously, A, B, and C get 4, 5, and 7 objects, respectively. Then number of distribution ways is equal to number of division of ways, which is given by 16!(4!5!7!).

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Correct Answer: C

Solution :

[c] Two circles intersect at two distinct points. Two straight lines intersect at one point. One circle and one straight line intersect at two distinct points. Then the total numbers of points of intersections are as follows;

Number of ways of selection	Points of intersection
Two straight lines: \[^{5}{{C}_{2}}\]	\[^{5}{{C}_{2}}\times 1=10\]
Two circles: \[^{4}{{C}_{2}}\]	\[^{4}{{C}_{2}}\times 2=12\]
One line and circle: \[^{5}{{C}_{1}}{{\times }^{4}}{{C}_{1}}\]	\[^{5}{{C}_{1}}{{\times }^{4}}{{C}_{1}}\times 2=40\]
Total	62

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Page 9

Correct Answer: D

Solution :
[d] Suppose \[{{i}^{th}}\]person receives Rs. \[{{x}_{i}};i=1,2,3,4\] Then, \[{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}=18,\]where \[{{x}_{i}}\ge 4\] Let \[{{y}_{i}}={{x}_{i}}-3,i=1,2,3,4.\]Then, \[{{y}_{1}}+{{y}_{2}}+{{y}_{3}}+{{y}_{4}}=6\] The total number of ways is equal to number of solutions of the above equation, which is given by \[^{6+4-1}{{C}_{4-1}}{{=}^{9}}{{C}_{3}}=84.\]

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Correct Answer: D

Solution :
[d] Let x, y, z be the friends and a, b, c denote the case when x is now, we have the following possibilities: (a, b, c) = (1, 2, 3) or (3 3 0) or (2 2 2) [grouping of 6 days of week] Hence, the total number of ways is \[\frac{6!}{1!2!3!}3!+\frac{6!}{3!3!2!}3!+\frac{6!}{(2!2!2!)}3!\]

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Correct Answer: B

Solution :
[b] There is concept of derangement. The require number is \[4!\left[ 1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!} \right]=9\]