Question

A delegation of four friends to be selected from a group of 12 friends. The number of ways the delegation will be selected if two particular friends refuse to be together and the two other particular friends wish to be together only in the delegation.
$\left( A \right)$ 226
$\left( B \right)$ 114
$\left( C \right)$ 156
$\left( D \right)$ 170

Answer 1

Hint – In this particular question use the concept that from the group of 12 friends separate 4 friends in which two wish to sit together and rest of the two does not wish to sit together then calculate the remaining friends in the group then create the possible cases of selecting 4 friends such that from these separated friends who wants to be together are in the delegation and who does not want to be together are not in the delegation so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given data:
There is a group of 12 friends.
From this group four friends were selected.
Now we have to find the number of ways that the delegation will be selected if two particular friends refuse to be together and two other particular friends wish to be together only in the delegation.
Now let these particular friends are A, B, C and D.
Where A and B wish to sit together and C and D does not wish to sit together.
So the remaining friends available in the group are (12 – 4) = 8.
Now the following cases are arises:
$\left( i \right)$ A, B, C is selected and D is not selected so we have to select one more person from the remaining persons in the group so that the delegation is completed.
So it is equal to ${}^8{C_1}$
$\left( {ii} \right)$ A, B, D is selected and C is not selected so we have to select one more person from the remaining persons in the group so that the delegation is completed.
So it is equal to ${}^8{C_1}$
$\left( {iii} \right)$ A, B is selected and C and D is not selected so we have to select two more people from the remaining persons in the group so that the delegation is completed.
So it is equal to ${}^8{C_2}$
$\left( {iv} \right)$ C is selected and A, B and D is not selected so we have to select three more people from the remaining persons in the group so that the delegation is completed.
So it is equal to ${}^8{C_3}$
$\left( v \right)$ D is selected and A, B and C is not selected so we have to select three more people from the remaining persons in the group so that the delegation is completed.
So it is equal to ${}^8{C_3}$
$\left( {vi} \right)$ None is selected, so we have to select four people from the remaining persons in the group so that the delegation is completed.
So it is equal to ${}^8{C_4}$
So the total number of ways to satisfy the given criteria is the sum of the all the cases so we have,
Number of cases = ${}^8{C_1} + {}^8{C_1} + {}^8{C_2} + {}^8{C_3} + {}^8{C_3} + {}^8{C_4}$
Now as we know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ so use this property in the above equation we have,
Number of cases = $\dfrac{{8!}}{{1!\left( {8 - 1} \right)!}} + \dfrac{{8!}}{{1!\left( {8 - 1} \right)!}} + \dfrac{{8!}}{{2!\left( {8 - 2} \right)!}} + \dfrac{{8!}}{{3!\left( {8 - 3} \right)!}} + \dfrac{{8!}}{{3!\left( {8 - 3} \right)!}} + \dfrac{{8!}}{{4!\left( {8 - 4} \right)!}}$
Now simplify this we have,
Number of cases = $\dfrac{{8!}}{{7!}} + \dfrac{{8!}}{{7!}} + \dfrac{{8!}}{{2!.6!}} + \dfrac{{8!}}{{3!.5!}} + \dfrac{{8!}}{{3!.5!}} + \dfrac{{8!}}{{4!.4!}}$
Number of cases = $8 + 8 + \dfrac{{8.7}}{2} + \dfrac{{8.7.6}}{{3.2.1}} + \dfrac{{8.7.6}}{{3.2.1}} + \dfrac{{8.7.6.5}}{{4.3.2.1}}$
Number of cases = $8 + 8 + 28 + 56 + 56 + 70 = 226$
So this is the required answer.

Hence option (A) is the correct answer.

Note – Whenever we face such types of questions the key concept we have to remember is that to select r object from the n objects we use combination rule (i.e. ${}^n{C_r}$) and also remember the formula of this combination which is stated above then the total number of ways of delegation of four friends is the sum of all of the above cases as above and simplify we will get the required answer.