Question

A woman has 11 close friends. Find the number of ways in which she can invite 5 of them to dinner if two particular persons are not on speaking terms and will not attend together.

Answer 1

Hint: We know that the total number of outcomes is the sum of the number of favorable outcomes and the number of unfavorable outcomes. Let us assume the total number of outcomes is equal to n(T), the total number of favorable outcomes is equal to n(F) and the total number of not favorable outcomes is equal to n(N). Let us consider this equation (1). Let us consider the total number of outcomes is the total number of ways to have 5 persons among 11 persons. Let us assume this as equation (2). Let us assume that the two particular persons attend the dinner as not favorable outcomes. So, the total number of not favorable outcomes is equal to the total number of ways that among 11 persons two persons must attend the meeting. Let us assume this as equation (3). So, now we have to find the number of favorable outcomes. The number of favorable outcomes is equal to the number of ways that two particular persons should attend the dinner. Now let by substituting equation (2) and equation (3) in equation (1). This will give us the value of n(F). Let us assume this as equation (4). So, in this way, we can get the number of ways in which she can invite 5 of them to dinner if two particular persons are not on speaking terms and will not attend together.

Complete step-by-step solution:
From the question, we were given that a woman has 11 friends and 5 must attend the dinner.
We know that the total number of outcomes is the sum of the number of favorable outcomes and the number of unfavorable outcomes.
Let us assume the total number of outcomes is equal to n(T), the total number of favorable outcomes is equal to n(F) and the total number of not favorable outcomes is equal to n(N).
Then we get
$$\Rightarrow n(T)=n(F)+n(N)....(1)$$
From the question, it is clear that
The total number of outcomes is the total number of ways to have 5 persons among 11 persons.
So, we get
$$\Rightarrow n(T){=}^{11}{C}_5.....(2)$$
Let us assume that the two particular persons attend the dinner as not favorable outcomes.
So, the total number of not favorable outcomes is equal to the total number of ways that among 11 persons two persons must attend the meeting.
If two persons should attend the meeting, then we have to choose another 3 persons among 9 other persons.
So, we get
$$\Rightarrow n(N){=}^9{C}_3......(3)$$
So, now we have to find the number of favorable outcomes.
The number of favorable outcomes is equal to the number of ways that two particular persons should attend the dinner.
Now we will substitute equation (2) and equation (3) in equation (1), then we get
$$\begin{array}{rl}& {\Rightarrow }^{11}{C}_5{=}^9{C}_3+n(F)\\ & \Rightarrow 462=84+n(F)\\ & \Rightarrow n(F)=378.....(4)\end{array}$$
From equation (4), it is clear that there are 378 ways in which she can invite 5 of them to dinner if two particular persons are not on speaking terms and will not attend together.

Note: This problem can be solved in an alternative manner.
Let us A and B will not attend the party together.
Case 1: (Let us assume A did not attend the party and B attend the party)
So, now we have to pick 4 persons from the remaining 9 persons.
The number of ways to have 4 persons among the 9 persons is equal to $${}^9{C}_4$$.
Case 2: (Let us assume B did not attend the party and A attend the party)
So, now we have to pick 4 persons from the remaining 9 persons.
The number of ways to have 4 persons among the 9 persons is equal to $${}^9{C}_4$$.
Case 3: (Let us assume both A and B did not attend the party)
So, now we have to pick 5 persons from the remaining 9 persons.
The number of ways to have 5 persons among the 9 persons is equal to $${}^9{C}_5$$.
So, let us assume the total number of ways is equal to T.
$$\begin{array}{rl}& \Rightarrow T{=}^9{C}_4{+}^9{C}_4{+}^9{C}_5\\ & \Rightarrow T=126+126+126\\ & \Rightarrow T=378....(1)\end{array}$$
So, from equation (1) it is clear that there are 378 ways in which she can invite 5 of them to dinner if two particular persons of them are not on speaking terms and will not attend together.