Question

A team of four students is to be selected from a total of $12$ students. The total number of ways in which the team can be selected such that two particular student refuse to be together and other two particular students wish to be together only is equal to
(a) $220$
(b) $182$
(c) $226$
(d) $495$

Answer 1

Hint: Let the two student which refuse to be together in the team be ${{S}_{1}}$ and ${{S}_{2}}$ . Also, let the student who wish together only be ${{S}_{3}}$ and ${{S}_{4}}$ . So, considered the possible different cases and then add the total number of ways in which they can occur. This will give us the total number of ways in which the team can be selected.

Complete step-by-step answer:

Here, we will use formula which says that when there are a set of n different objects out of which r object or to be selected, then number of ways to select \[={{n}_{{{C}_{r}}}}=\dfrac{n!}{r!\left( n-r \right)!}\]
Case – ${{1}^{st}}$ : We take ${{S}_{1}}$ in the team. In this case ${{S}_{2}}$ is not taken.
 Subcase 1- a: After taking ${{S}_{1}}$ in team we take ${{S}_{3}}$ and ${{S}_{4}}$ in the team. So, now, only one place remains vacant for other student to take. Now, since ${{S}_{1}}$ is included along with ${{S}_{3}}$ and ${{S}_{4}}$ so, ${{S}_{2}}$ cannot be included. So, remains $12-4=8$ students who can participate in the team out of which only one is to be selected.
So, number of ways to do so is we have to select one student out of eight different students $={}^{8}{{C}_{1}}$
 So, number ways to select in case (1- a ) $={}^{8}{{C}_{1}}=8$ $\left( \because {}^{8}{{C}_{1}}=\dfrac{8!}{\left( 8-1 \right)!\cdot 1!}=\dfrac{8\times 7!}{7!}=8 \right)$ ……(1)
 Subcase 1 – b: In this case ${{S}_{1}}$ is selected (so, ${{S}_{2}}$ isn’t selected) but ${{S}_{3}}$ and ${{S}_{4}}$ aren’t selected. So, total vacant place remained in the team $=4-1=3$ and total student who can be selected are $12-4=8$ . So, we have to select 3 students from the remaining 8, we get
$={{8}_{{{C}_{3}}}}=\dfrac{8!}{3!\left( 8-3 \right)!}=\dfrac{8\times 7\times 6\times 5!}{3\times 2\times 1\times 5!}=8\times 7=56$ …………………..(2)
Case - ${{2}^{nd}}$ : When ${{S}_{1}}$ isn’t selected and ${{S}_{2}}$ is selected.
Subcase 2-a: In this case ${{S}_{2}}$ is selected (so, ${{S}_{1}}$ not selected) along with ${{S}_{3}}$ and ${{S}_{4}}$ .So, three students selected. Hence, one place remains vacant for selection from $\left( 12-4 \right)$ i.e. $\left( {{S}_{1}}\text{, }{{S}_{2}},\text{ }{{S}_{3}},\text{ }{{S}_{4}} \right)=8$
So, number of ways to do it $={}^{8}{{C}_{1}}=8$ …………………(3)
Subcase 2 -b: In this case ${{S}_{2}}$ selected so, ${{S}_{1}}$ not selected. Also, ${{S}_{3}}$ and ${{S}_{4}}$ are not selected. So, $4-1=3$ seats remain vacant for selection from $12-4=8$ students.
So, number of ways to select $={{8}_{{{C}_{3}}}}=\dfrac{8!}{3!\left( 8-3 \right)!}=\dfrac{8\times 7\times 6\times 5!}{3\times 2\times 1\times 5!}=8\times 7=56$ ……….(4)
Case - ${{3}^{rd}}$ : In this case ${{S}_{1}}$ and ${{S}_{2}}$ are not selected .
Subcase 3 -a: In this case ${{S}_{3}}$ and ${{S}_{4}}$ are selected. So, two seats remained in team. Now, since ${{S}_{1}}$ and ${{S}_{2}}$ are not option to be selected and ${{S}_{3}},\text{ }{{S}_{4}}$ are selected. So, total $12-4=8$ students are remained from which the seats can be filled.
So, number of ways to do $={}^{8}{{C}_{2}}=\dfrac{8!}{2!\left( 8-2 \right)!}=\dfrac{8\times 7\times 6!}{2\times 6!}=4\times 7=28$ …………………(5)
Subcase 3 – b: ${{S}_{1}}$ and ${{S}_{2}}$ are rejected. Also, ${{S}_{3}}$ and ${{S}_{4}}$ are rejected.
So, in this case four students are not counted for selection $\left( {{S}_{1}},\text{ }{{S}_{2}},\text{ }{{S}_{3}},\text{ }{{S}_{4}} \right)$ . So, available students are remaining 8. So, number of ways to select four members in team
$={}^{8}{{C}_{4}}=\dfrac{8!}{4!\left( 8-4 \right)!}=\dfrac{8\times 7\times 6\times 5\times 4!}{4\times 3\times 2\times 1\times 4!}=7\times 5\times 2=70$ ……………………….(6)
Now, total number of ways to select the team is the addition of number of ways in case first, second and third i.e. addition of equation (1), (2), (3), (4), (5), (6) .
Therefore, total number of ways in which the team can be selected will be
$\Rightarrow \left[ {}^{8}{{C}_{1}}+{}^{8}{{C}_{3}} \right]+\left[ {}^{8}{{C}_{1}}+{}^{8}{{C}_{3}} \right]+{}^{8}{{C}_{2}}+{}^{8}{{C}_{4}}$
$\Rightarrow 2\left[ {}^{8}{{C}_{1}}+{}^{8}{{C}_{3}} \right]+{}^{8}{{C}_{2}}+{}^{8}{{C}_{4}}$
$\Rightarrow 2\left[ \dfrac{8!}{\left( 8-1 \right)!\cdot 1!}+\dfrac{8!}{\left( 8-3 \right)!\cdot 3!} \right]+\dfrac{8!}{\left( 8-2 \right)!\cdot 2!}+\dfrac{8!}{\left( 8-4 \right)!\cdot 4!}$
$\Rightarrow 2\left[ \dfrac{8\times 7!}{7!}+\dfrac{8\times 7\times 6\times 5!}{5!\times 3\times 2\times 1} \right]+\dfrac{8\times 7\times 6!}{6!\cdot 2\times 1}+\dfrac{8\times 7\times 6\times 5!}{4!\cdot 4\times 3\times 2\times 1}$
$\Rightarrow 2\left( 8+56 \right)+28+70$
$\Rightarrow 2\left( 64 \right)+28+70$
$\Rightarrow 226$
Hence, we can select the team by following the condition given in 226 ways.
So, correct option is option (C).

Note: Alternatively one may also calculate total number of ways by selecting student from 12 students and subtracting the cases in which ${{S}_{1}}$ and ${{S}_{2}}$ are together and ${{S}_{3}}$ and ${{S}_{4}}$ are not together. But that will give more number of cases which should be avoided.