Question

Six married couples are sitting in a room. Find the number of ways in which 4 people can be selected so that
(a) they do not form a couple
(b) they form exactly one couple
(c) they form at least one couple
(d) they form at most one couple

Answer 1

Hint: Here, the uniform rate is given so we can use simple methods of proportion to solve this problem. Use the fact that if two things are in direct proportion, then the rate of increasing or decreasing of one is the same as the rate of increasing or decreasing of the other.

Complete Step-by-step Solution
(a) they do not form a couple
In this part, it is given that 4 people can be selected so that they don’t form a couple. Here, the people are not forming a couple and we will find the number of ways for the same.
If four people selected do not form a couple that means that they were selected from four different couples.
Ways of selecting four couple from six is,
${}^6{C_4} = \dfrac{{6!}}{{4!\left( {6 - 4} \right)!}} = 15$
Ways of selecting four persons from each couple is ${}^2{C_1} \times {}^2{C_1} \times {}^2{C_1} \times {}^2{C_1}$,
${}^2{C_1} \times {}^2{C_1} \times {}^2{C_1} \times {}^2{C_1} = 16$
So, total ways of selecting 4 persons so that they do not form couple is,
$15 \times 16 = 240$
Therefore, the number of ways in which 4 people can be selected is $240$.

(b) they form exactly one couple
In this part, it is given that 4 people can be selected in which they form exactly one couple. Here, in 4 people there is one couple. So, we will find the number of ways for the same.
If four people are selected from exactly one couple then they are selected from 3 couples (2 from first couple, 1 from second couple, 1 from third couple).
Ways of selecting three couple from six is,
${}^6{C_3} = \dfrac{{6!}}{{3!\left( {6 - 3} \right)!}} = 20$
Ways of selecting 1 couple from three couple is,
${}^3{C_1} = 3$
Ways of selecting two from the selected one couple and one each from remaining two couple is,${}^2{C_2} \times {}^2{C_1} \times {}^2{C_1} = 4$
So, total ways of selecting 4 persons so that they form exactly one couple is,
$20 \times 3 \times 4 = 240$
Therefore, the number of ways in which 4 people can be selected is $240$.

(c) they form at least one couple
In this part, it is given that 4 people can be selected in which they form at least one couple. This means that 1 couple will definitely be formed. So, we will find the number of ways for the same.
Total ways of selecting 4 people from 12-Ways of selecting so that they do not form couple
\[\begin{array}{c}
{}^{12}{C_4} - 240 = {\rm{ }}495 - 240\\
 = {\rm{ }}255
\end{array}\]
Therefore, the number of ways in which 4 people can be selected is $255$.

(d) they form at most one couple
In this part, it is given that 4 people can be selected in which they form at most one couple. This means that two couples will definitely be formed. So, we will find the number of ways for the same.
 Ways such that they do not form a couple + Ways such that they form exactly one couple is,
\[240 + 240 = {\rm{ }}480\]
Therefore, the number of ways in which 4 people can be selected is $480$.


Note: General formula for combination is ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ that is these are the number of ways in which r things can be selected out of n things. In part (c) and (d) students might go wrong because at least means not less than and at most not more than are the terms which might create confusion.