Q.

A boy puts his hand into a bag which contains 10 differently coloured marbles and brings out 3. How many different results are possible?

➤

Total number of differently coloured marbles = 10
Number of marbles drawn = 3
Number of different results possible = \(^{10}C_3\) = \(\frac{10!}{3!7!}\) = 120

Q.

Find the number of ways in which a student can select 5 courses out of 8 courses. If 3 courses are compusory, in how many ways can the selections be made?

➤

Total number of courses = 8
Number of courses to be selected = 5
Number of ways in which a student can select 5 courses out of 8 courses = \(^{8}C_5\) = \(\frac{8!}{5!3!}\) = 56
Number of compulsory courses = 3
Number of remaining courses to be selected = 2
Number of ways in which the selections can be made = \(^{5}C_2\) = \(\frac{5!}{2!3!}\) = 10

Q.

From 10 persons, in how many ways can a selection of 4 be made

1. when one particular person is always included?
2. when two particular persons are always excluded?

➤

Total number of persons = 10
Number of persons to be selected = 4

1. Number of ways in which a selection of 4 can be made when one particular person is always included = \(^{9}C_3\) = \(\frac{9!}{3!6!}\) = 84
2. Number of ways in which a selection of 4 can be made when two particular persons are always excluded = \(^{8}C_4\) = \(\frac{8!}{4!4!}\) = 70
    

Q.

A bag contains 8 white balls and 5 blue balls. In how many ways can 5 white balls and 3 blue balls be drawn?

➤

Number of white balls = 8
Number of blue balls = 5
Number of white balls to be drawn = 5
Number of blue balls to be drawn = 3
Number of ways in which 5 white balls and 3 blue balls can be drawn = \(^{8}C_5 \times ^{5}C_3\) = \(\frac{8!}{5!3!} \times \frac{5!}{3!2!} = 56 \times 10 = 560\)
 

Q.

How many committees can be formed from a set of 7 boys and 5 girls if each committee contains 4 boys and 3 girls?

➤

For boys committee:
Number of boys = 7
Number of boys to be selected = 4
Number of boys committees = \(^{7}C_4\) = \(\frac{7!}{4!3!}\) = 35

For girls committee:
Number of girls = 5
Number of girls to be selected = 3
Number of girls committees = \(^{5}C_3\) = \(\frac{5!}{3!2!}\) = 10

Total number of committees = \(35 \times 10 = 350 \)

Q.

From a group of 11 men and 8 women, how many committees consisting of 3 men and 2 womens are possible?

➤

For men's committee:
Total number of men = 11
Number of men to be selected = 3
Number of ways to select 3 men from 11 = \(^{11}C_3\) = \(\frac{11!}{3!8!}\) = 165

For women's committee:
Total number of women = 8
Number of women to be selected = 2
Number of ways to select 2 women from 8 = \(^{8}C_2\) = \(\frac{8!}{2!6!}\) = 28

Total number of committees = \(165 \times 28 = 4620\)

Q.

From 4 mathematicians, 6 stratisticians and 5 economists, how many committees of 6 members can be formed so as to include 2 members from each category?

➤

For mathematicians:
Total number of mathematicians = 4
Number of mathematicians to be selected = 2
Number of ways to select 2 mathematicians from 4 = \(^{4}C_2\) = \(\frac{4!}{2!2!}\) = 6

For statisticians:
Total number of statisticians = 6
Number of statisticians to be selected = 2
Number of ways to select 2 statisticians from 6 = \(^{6}C_2\) = \(\frac{6!}{2!4!}\) = 15

For economists:
Total number of economists = 5
Number of economists to be selected = 2
Number of ways to select 2 economists from 5 = \(^{5}C_2\) = \(\frac{5!}{2!3!}\) = 10

Total number of committees = \(6 \times 15 \times 10 = 900\)
\end{solution}
 

Q.

A person has got 12 acquaintances of whom 8 are relatives. In how many ways can he invite 7 guests so that 5 of them may be relatives?

➤

Total number of acquaintances = 12
Number of relatives = 8
Number of non-relatives = 4
Number of guests to be invited = 7
Number of relatives to be invited = 5
Number of non-relatives to be invited = 2

Number of ways to select 5 relatives from 8 = \(^{8}C_5\) = \(\frac{8!}{5!3!}\) = 56
Number of ways to select 2 non-relatives from 4 = \(^{4}C_2\) = \(\frac{4!}{2!2!}\) = 6

Total number of ways to invite 7 guests so that 5 of them are relatives = \(56 \times 6 = 336\)
 

Q.

There are ten electric bulbs in the stock of a shop out of which there are three defectives. In how many ways can a selection of 6 bulbs be made so that 4 of them may be good bulbs?

➤

Total number of bulbs = 10
Number of bulbs to be selected = 6

For non-defective bulbs: 
Number of non-defective bulbs = 7
Number of good bulbs to be selected = 4
Number of ways to select 4 good bulbs from 7 = \(^{7}C_4\) = \(\frac{7!}{4!3!}\) = 35

For defective bulbs: 
Number of defective bulbs = 3
Number of defective bulbs to be selected = 2
Number of ways to select 2 defective bulbs from 3 = \(^{3}C_2\) = \(\frac{3!}{2!1!}\) = 3

Total number of ways to select 6 bulbs so that 4 of them are good = \(35 \times 3 = 105\)
 

Q.

From 6 gentleman and 4 ladies, a committee of 5 is to be formed. In how many ways can this be done so as to include at least one lady?

➤

Total number of gentlemen = 6
Total number of ladies = 4
Number of members to be selected = 5

Number of ways to select 1 lady and 4 gentlemen = \(^{4}C_1 \times ^{6}C_4\) = \(\frac{4!}{1!3!} \times \frac{6!}{4!2!}\) = 60
Number of ways to select 2 ladies and 3 gentlemen = \(^{4}C_2 \times ^{6}C_3\) = \(\frac{4!}{2!2!} \times \frac{6!}{3!3!}\) = 120
Number of ways to select 3 ladies and 2 gentlemen = \(^{4}C_3 \times ^{6}C_2\) = \(\frac{4!}{3!1!} \times \frac{6!}{2!4!}\) = 60
Number of ways to select 4 ladies and 1 gentleman = \(^{4}C_4 \times ^{6}C_1\) = \(\frac{4!}{4!0!} \times \frac{6!}{1!5!}\) = 6

Total number of ways to form a committee of 5 members so as to include at least one lady = \(60 + 120 + 60 + 6 = 246\) ways
 

Q.

A candidate is required to answer 6 out of 10 questions which are divided into 2 groups each containing 5 questions and he is not permitted to attempt more than 4 from any group. In how many different ways can he make up his choice?

➤

Given:
Total number of questions = 10
Number of questions in each group = 5
Number of questions to be answered = 6
Maximum questions from each group = 4

We need to find the number of ways to select 6 questions such that no more than 4 are from any group.

Case 1: Selecting 2 questions from group A and 4 questions from group B
\[
\text{Number of ways to select 2 questions from group A} = ^{5}C_2 = \frac{5!}{2!3!} = 10
\]
\[
\text{Number of ways to select 4 questions from group B} = ^{5}C_4 = \frac{5!}{4!1!} = 5
\]
\[
\text{Total number of ways for this case} = 10 \times 5 = 50
\]

Case 2: Selecting 3 questions from group A and 3 questions from group B
\[
\text{Number of ways to select 3 questions from group A} = ^{5}C_3 = \frac{5!}{3!2!} = 10
\]
\[
\text{Number of ways to select 3 questions from group B} = ^{5}C_3 = \frac{5!}{3!2!} = 10
\]
\[
\text{Total number of ways for this case} = 10 \times 10 = 100
\]

Case 3: Selecting 4 questions from group A and 2 questions from group B
\[
\text{Number of ways to select 4 questions from group A} = ^{5}C_4 = \frac{5!}{4!1!} = 5
\]
\[
\text{Number of ways to select 2 questions from group B} = ^{5}C_2 = \frac{5!}{2!3!} = 10
\]
\[
\text{Total number of ways for this case} = 5 \times 10 = 50
\]

Summing all the cases:
\[
\text{Total number of ways to select 6 questions} = 50 + 100 + 50 = 200
\]
 

Q.

A man has 5 friends. In how many ways can he invite one or more of them to a dinner?

➤

Given:
Number of friends = 5
Number of friends to be invited = 1 or more

Number of ways to invite 1 friend = \(^{5}C_1\) = \(\frac{5!}{1!4!}\) = 5
Number of ways to invite 2 friends = \(^{5}C_2\) = \(\frac{5!}{2!3!}\) = 10
Number of ways to invite 3 friends = \(^{5}C_3\) = \(\frac{5!}{3!2!}\) = 10
Number of ways to invite 4 friends = \(^{5}C_4\) = \(\frac{5!}{4!1!}\) = 5
Number of ways to invite 5 friends = \(^{5}C_5\) = \(\frac{5!}{5!0!}\) = 1

Total number of ways to invite one or more friends = 5 + 10 + 10 + 5 + 1 = 31
 

Q.

1. If \(C(20, r+5) = C(20, 2r-7)\), find \(C(15, r)\).
2. If \(C(n, 10) + C(n, 9) = C(20, 10)\), find n and \(C(n, 17)\).
3. Solve for n the equation \(C(n+2, 4) = 6 C(n, 2)\).
4. If \(P(n,r) = 336 \) and \(C(n, r) = 56\), find n and r.
5. If \(nC_(r+1) = 45, nC_r = 120 \) and \(nC_(r+1) = 210 \), find n and r.

➤

1. Since the combinations are equal, we can equate their r components:
                \begin{align}
                    r + 5 & = 2r - 7 \\
                    5 + 7 & = 2r - r \\
                    r     & = 12
                \end{align}
   Now, we need to find \(C(15, r)\):
                \begin{align}
                    C(15, 12) & = C(15, 3)                                           \\
                              & = \frac{15!}{3!(15-3)!}                              \\
                              & = \frac{15!}{3!12!}                                  \\
                              & = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} \\
                              & = 455
                \end{align}
                So, \(C(15, 12) = 455\).
2. Given:
                \begin{align}
                    C(n, 10) + C(n, 9)                                            & = C(20, 10)               \\
                    \frac{n!}{10!(n-10)!} + \frac{n!}{9!(n-9)!}                   & = \frac{20!}{10!(20-10)!} \\
                    \frac{n!}{10!(n-10)!} + \frac{n!}{9!(n-9)!}                   & = \frac{20!}{10!10!}      \\
                    \frac{n!}{10!(n-10)!} + \frac{n! \cdot 10}{10 \cdot 9!(n-9)!} & = \frac{20!}{10!10!}      \\
                    \frac{n!}{10!(n-10)!} + \frac{10 \cdot n!}{10!(n-9)!}         & = \frac{20!}{10!10!}      \\
                    \frac{n!}{(n-10)!} + \frac{10 \cdot n!}{(n-9)!}               & = \frac{20!}{10!}         \\
                    \frac{n!(n-9)}{(n-10)!(n-9)} + \frac{10 \cdot n!}{(n-9)!}     & = \frac{20!}{10!}         \\
                    \frac{n!(n-9)}{(n-9)!} + \frac{10 \cdot n!}{(n-9)!}           & = \frac{20!}{10!}         \\
                    \frac{n!}{(n-9)!}(n-9+10)                                     & = \frac{20!}{10!}         \\
                    \frac{n!}{(n-9)!}(n+1)                                        & = \frac{20!}{10!}         \\
                    \frac{(n+1)!}{(n-9)!}                                         & = \frac{20!}{10!}         \\
                \end{align}

             Solving for \(n\):
             \begin{align}
                 (n+1)! & = 20! \\
                 n+1    & = 20  \\
                 n      & = 19
             \end{align}
             Now, we need to find \(C(n, 17)\):
             \begin{align}
                 C(19, 17) & = C(19, 2)                        \\
                           & = \frac{19!}{2!(19-2)!}           \\
                           & = \frac{19!}{2!17!}               \\
                           & = \frac{19 \times 18}{2 \times 1} \\
                           & = 171
             \end{align}
             So, \(C(19, 17) = 171\).

c.                 \begin{align}
                 C(n+2, 4)                 & = 6 C(n, 2)             \\
                 \frac{(n+2)!}{4!(n+2-4)!} & = 6 \frac{n!}{2!(n-2)!} \\
                 \frac{(n+2)!}{4!(n-2)!}   & = 3 \frac{n!}{(n-2)!}   \\
                 \frac{(n+2)!}{4!}         & = 3 \cdot n!            \\
                 \frac{(n+2)(n+1)n!}{24}   & = 3n!                   \\
                 \frac{(n+2)(n+1)}{24}     & = 3                     \\
                 (n+2)(n+1)                & = 72                    \\
                 n^2 + 3n + 2              & = 72                    \\
                 n^2 + 3n - 70             & = 0
             \end{align}
             Using the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):   \[
                 n = \frac{-3 \pm \sqrt{9 + 280}}{2} = \frac{-3 \pm 17}{2}
             \]
             So, \(n = 7\) or \(n = -10\). Since \(n\) must be non-negative, \(n = 7\).

d.   Given:
             \[
                 P(n, r)  = 336  \text{ and }  C(n, r) = 56
             \]

             We know that:
             \begin{align}
                 P(n, r) & = r! \cdot C(n, r)   \\
                 336     & = r! \cdot 56        \\
                 r!      & = \frac{336}{56} = 6 \\
                 r       & = 3
             \end{align}

             Now, we need to find \(n\) using \(P(n, r) = 336\):
             \begin{align}
                 P(n, 3)           & = \frac{n!}{3!(n-3)!} = 336                                    \\
                 \frac{n!}{(n-3)!} & = 336                                                          \\
                 \frac{8!}{(8-3)!} & = \frac{40320}{120} = 336 \quad \text{(Trying with \(n = 8\))}
             \end{align}
             So, \(n = 8\) and \(r = 3\).

e.   Given:
             \[
                 C(n, r-1) = \frac{n!}{(r-1)!(n-r+1)!} = 45 \tag{1}
             \]
             \[
                 C(n, r) = \frac{n!}{r!(n-r)!} = 120 \tag{2}
             \]
             \[
                 C(n, r+1) = \frac{n!}{(r+1)!(n-r-1)!} = 210 \tag{3}
             \]

             Dividing equation (2) by equation (1):
             \begin{align}
                 \frac{C(n, r)}{C(n, r-1)}                             & = \frac{120}{45}               \\
                 \frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r-1)!(n-r+1)!}} & = \frac{120}{45}               \\
                 \frac{(n-r+1)}{r}                                     & = \frac{120}{45} = \frac{8}{3} \\
                 3(n-r+1)                                              & = 8r                           \\
                 3n - 3r + 3                                           & = 8r                           \\
                 3n + 3                                                & = 11r                          \\
                 r                                                     & = \frac{3n + 3}{11}
             \end{align}

             Dividing equation (3) by equation (2):
             \begin{align}
                 \frac{C(n, r+1)}{C(n, r)}                             & = \frac{210}{120}               \\
                 \frac{\frac{n!}{(r+1)!(n-r-1)!}}{\frac{n!}{r!(n-r)!}} & = \frac{210}{120}               \\
                 \frac{(n-r)}{(r+1)}                                   & = \frac{210}{120} = \frac{7}{4} \\
                 4(n-r)                                                & = 7(r+1)                        \\
                 4n - 4r                                               & = 7r + 7                        \\
                 4n                                                    & = 11r + 7                       \\
                 r                                                     & = \frac{4n - 7}{11}
             \end{align}

             Equating the two expressions for \(r\):
             \begin{align}
                 \frac{3n + 3}{11} & = \frac{4n - 7}{11} \\
                 3n + 3            & = 4n - 7            \\
                 n                 & = 10
             \end{align}

             Substituting \(n = 10\) back into one of the equations for \(r\):
             \[
                 r = \frac{3(10) + 3}{11} = \frac{30 + 3}{11} = 3
             \]
             So, \(n = 10\) and \(r = 3\).

Q.

An examination paper consisting of 10 questions, is divided into two groups A and B. Group A contains 6 questions. In how many ways can an examinee attempt 7 questions.

1. selecting 4 from group A and 3 from group B?
2. selecting atleast two questions from each group?

➤

    (i) Selecting 4 from group A and 3 from group B:
   Number of ways to select 4 questions from group A = \(^{6}C_4 = \frac{6!}{4!2!} = 15\)
   Number of ways to select 3 questions from group B = \(^{4}C_3 = \frac{4!}{3!1!} = 4\)
   Total number of ways for this case = \(15 \times 4 = 60\)

   (ii) Selecting at least two questions from each group:
   Case 1: Selecting 2 questions from group A and 5 questions from group B
   Number of ways to select 2 questions from group A = \(^{6}C_2 = \frac{6!}{2!4!} = 15\)
   Number of ways to select 5 questions from group B = \(^{4}C_5 = 0\) (not possible)
   Total number of ways for this case = 0

   Case 2: Selecting 3 questions from group A and 4 questions from group B
   Number of ways to select 3 questions from group A = \(^{6}C_3 = \frac{6!}{3!3!} = 20\)
   Number of ways to select 4 questions from group B = \(^{4}C_4 = 1\)
   Total number of ways for this case = \(20 \times 1 = 20\)

   Case 3: Selecting 4 questions from group A and 3 questions from group B
   Number of ways to select 4 questions from group A = \(^{6}C_4 = \frac{6!}{4!2!} = 15\)
   Number of ways to select 3 questions from group B = \(^{4}C_3 = \frac{4!}{3!1!} = 4\)
   Total number of ways for this case = \(15 \times 4 = 60\)

   Case 4: Selecting 5 questions from group A and 2 questions from group B
   Number of ways to select 5 questions from group A = \(^{6}C_5 = \frac{6!}{5!1!} = 6\)
   Number of ways to select 2 questions from group B = \(^{4}C_2 = \frac{4!}{2!2!} = 6\)
   Total number of ways for this case = \(6 \times 6 = 36\)

   Summing all the cases:
   Total number of ways to select 7 questions = \(0 + 20 + 60 + 36 = 116\)

Q.

Six men in a group of 8 are skilled. Find the number of ways by which 5 men can be slected such that:

1. atleast 3 of them may be the skilled man.
2. atleast one of them may be the unskilled man.

➤

  (i) At least 3 of them may be skilled men:
   Case 1: Selecting 3 skilled men and 2 unskilled men:
   Number of ways to select 3 skilled men from 6 = \(^{6}C_3 = \frac{6!}{3!3!} = 20\)
   Number of ways to select 2 unskilled men from 2 = \(^{2}C_2 = \frac{2!}{2!0!} = 1\)
   Total number of ways for this case = \(20 \times 1 = 20\)

   Case 2: Selecting 4 skilled men and 1 unskilled man:
   Number of ways to select 4 skilled men from 6 = \(^{6}C_4 = \frac{6!}{4!2!} = 15\)
   Number of ways to select 1 unskilled man from 2 = \(^{2}C_1 = \frac{2!}{1!1!} = 2\)
   Total number of ways for this case = \(15 \times 2 = 30\)

   Case 3: Selecting 5 skilled men:
   Number of ways to select 5 skilled men from 6 = \(^{6}C_5 = \frac{6!}{5!1!} = 6\)

   Summing all the cases:
   Total number of ways to select 5 men such that at least 3 are skilled = 20 + 30 + 6 = 56

   (ii) At least one of them may be the unskilled man:
   Total number of ways to select 5 men from 8 = \(^{8}C_5 = \frac{8!}{5!3!} = 56\)

   Number of ways to select 5 skilled men from 6 = \(^{6}C_5 = \frac{6!}{5!1!} = 6\)

   Total number of ways to select 5 men such that at least one is unskilled = 56 - 6 = 50

Q.

In a group of 10 students, 6 are boys. In how many ways can 4 students are slected for mathematical competitions so as to include

1. exactly two boys
2. atleast two boys
3. at most two girls

➤

  (i) At least two boys:
   Case 1: Selecting 2 boys and 2 girls:
   Number of ways to select 2 boys from 6 = \(^{6}C_2 = \frac{6!}{2!4!} = 15\)
   Number of ways to select 2 girls from 4 = \(^{4}C_2 = \frac{4!}{2!2!} = 6\)
   Total number of ways for this case = \(15 \times 6 = 90\)

   Case 2: Selecting 3 boys and 1 girl:
   Number of ways to select 3 boys from 6 = \(^{6}C_3 = \frac{6!}{3!3!} = 20\)
   Number of ways to select 1 girl from 4 = \(^{4}C_1 = \frac{4!}{1!3!} = 4\)
   Total number of ways for this case = \(20 \times 4 = 80\)

   Case 3: Selecting 4 boys:
   Number of ways to select 4 boys from 6 = \(^{6}C_4 = \frac{6!}{4!2!} = 15\)

   Summing all the cases:
   Total number of ways to select 4 students such that at least two are boys = \(90 + 80 + 15 = 185\)

   (ii) At most two girls:
   Case 1: Selecting 0 girls and 4 boys:
   Number of ways to select 4 boys from 6 = \(^{6}C_4 = \frac{6!}{4!2!} = 15\)

   Case 2: Selecting 1 girl and 3 boys:
   Number of ways to select 1 girl from 4 = \(^{4}C_1 = \frac{4!}{1!3!} = 4\)
   Number of ways to select 3 boys from 6 = \(^{6}C_3 = \frac{6!}{3!3!} = 20\)
   Total number of ways for this case = \(4 \times 20 = 80\)

   Case 3: Selecting 2 girls and 2 boys:
   Number of ways to select 2 girls from 4 = \(^{4}C_2 = \frac{4!}{2!2!} = 6\)
   Number of ways to select 2 boys from 6 = \(^{6}C_2 = \frac{6!}{2!4!} = 15\)
   Total number of ways for this case = \(6 \times 15 = 90\)

   Summing all the cases:
   Total number of ways to select 4 students such that at most two are girls = (15 + 80 + 90 = 185)