Question

There are $10$ girls and $8$ boys in a class room including Mr. Ravi, Ms. Rani and Ms. Radha. A list of speakers consisting of $8$ girls and $6$ boys has to be prepared. Mr. Ravi refuses to speak if Ms. Rani is a speaker. Ms. Rani refuses to speak if Ms. Radha is a speaker. The number of ways the list can be prepared is a 3-digit number ${n_1}{n_2}{n_3}$, then$\left| {{n_3} + {n_2} - {n_1}} \right| = ?$

Answer 1

In case, there were no personality conflicts among the speakers we would have taken two combinations (one for the girls and one for the boys) and then multiplied them both to get the total number of ways the list can be prepared.

{}^{10}{C_8} \times {}^8{C_6} = \dfrac{{10!}}{{\left( {8!} \right)\left( {2!} \right)}} \times \dfrac{{8!}}{{\left( {6!} \right)\left( {2!} \right)}} \\\ = \dfrac{{10!}}{{\left( {6!} \right){{\left( {2!} \right)}^2}}} \\\ = 1260 $$ If there were no conflicts, there would have been $$1260$$ ways to prepare the list. But, in the given question there are personality conflicts and it is given that the total number the list can be prepared is a three digit number. Largest three digit number is $$999$$. In any case, the number of ways can be greater than $$999$$. Formula used: $${}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}},n \geqslant r$$ **Complete step-by-step solution:** To solve the question, we will consider each case separately. We will find the acceptable number of ways the list can be prepared after considering the given conditions and sum it all up at the end. Limitations arise when Ms. Rani or Ms. Radha speaks and so we will have three cases- Ms. Radha is a speaker, If Ms. Radha is not speaker, two subcases will be there-Ms. Rani is a speaker and Ms. Rani is not a speaker. CASE 1: When Ms. Radha is a speaker If Ms. Radha speaks, Ms. Rani would not speak. In that case, we will select $$7$$ girls out of remaining $$8$$ girls and $$6$$ boys from $$8$$ boys. The number of ways can be:

{}^8{C_7} \times {}^8{C_6} = \dfrac{{8!}}{{\left( {7!} \right)\left( {1!} \right)}} \times \dfrac{{8!}}{{\left( {6!} \right)\left( {2!} \right)}} \\
\\

$$$$

= 8 \times 28 \\
= 224 \\

CASE 2: When Ms. Radha is not a speaker, but Ms. Rani is a speaker and Mr. ravi would not speak in this case. Hence In this case, we will select $$7$$ girls out of remaining $$8$$ girls and $$6$$ boys from remaining $$7$$ boys which can be done in:

{}^8{C_7} \times {}^7{C_6} = \dfrac{{8!}}{{\left( {7!} \right)\left( {1!} \right)}} \times \dfrac{{7!}}{{\left( {6!} \right)\left( {1!} \right)}} \\
= 8 \times 7 \\

$$ = 56$$ways. CASE 3: Neither Ms. Radha nor Ms. Rani is a speaker In this case, we will select $$8$$ girls out of remaining $$8$$ girls and $$6$$ boys from $$8$$ boys which can be done in:

{}^8{C_6} \times {}^8{C_8} = \dfrac{{8!}}{{\left( {6!} \right)\left( {2!} \right)}} \times \dfrac{{8!}}{{\left( {8!} \right)\left( {0!} \right)}} \\
= 28 \times 1 \\

$$ = 28$$ways Hence, the total number of ways the list can be prepared given the conflicts is $$224 + 56 + 28 = 308$$ ways. Now, $$308$$ is a three digit number with $${n_1} = 3,{n_2} = 0,{n_3} = 8$$ Then, $$\left| {{n_3} + {n_2} - {n_1}} \right| = \left| {8 + 0 - 3} \right|$$ $$ \Rightarrow \left| {{n_3} + {n_2} - {n_1}} \right| = 5$$ **Note:** While doing such types of questions, we should not be in a hurry, usually we have a habit of solving Permutations and Combinations and this may lead us to leave the solution halfway. Read the question carefully and then solve for what is asked in it. It is very important to have thorough knowledge of the concept of Combination before solving such types of questions. A minor mistake will give a wrong result.