A candidate is required to answer 7 questions, out of 12 questions which are divided into two groups, each containing 6 questions. He is not permitted to attempt more than 5 questions from either group. He can choose the seven questions in 650 ways.
To fill 12 vacancies there are 25 candidates of which 5 are from scheduled castes. If 3 of the vacancies are reserved for scheduled caste candidates while the rest are open to all, the number of ways in which the selection can be made is ${ }^5 C_3 \times{ }^{20} C_9$.
There are 3 books on Mathematics, 4 on Physics and 5 on English. How many different collections can be made such that each collection consists?
Column I | Column II | ||
---|---|---|---|
(i) | One book of each subject | (a) | 3968 |
(ii) | Atleast one book of each subject | (b) | 60 |
(iii) | Atleast one book of English | (c) | 3255 |
There are three books of Mathematics 4 of Physics and 5 on English.
$$\begin{aligned} \text { (i) One book of each subject } & ={ }^3 C_1 \times{ }^4 C_1 \times{ }^5 C_1 \\ & =3 \times 4 \times 5=60 \end{aligned}$$
(ii) Atleast one book of each subject $=\left(2^3-1\right) \times\left(2^4-1\right) \times\left(2^5-1\right)$
$$=7 \times 15 \times 31=3255$$
(iii) Atleast one book of English = Selection based on following manner
English book | 1 | 2 | 3 | 4 | 5 |
---|---|---|---|---|---|
Others | 11 | 10 | 9 | 8 | 7 |
$\begin{aligned} & =\left(2^5-1\right) \times 2^7 \\ & =128 \times 31=3968\end{aligned}$
Five boys and five girls form a line. Find the number of ways of making the seating arrangement under the following condition.
Column I | Column II | ||
---|---|---|---|
(i) | Boys and girls alternate | (a) | $5!\times6!$ |
(ii) | No two girls sit together | (b) | $10!-5! 6!$ |
(iii) | All the girls sit together | (c) | $(5!)^2+(5!)^2$ |
(iv) | All the girls are never together | (d) | $2! 5! 5!$ |
(i) Boys and girls alternate
Total arrangements $=(5!)^2+(5!)^2$
(ii) No two girls sit together $=5 ! 6 !$
(iii) All the girls sit together $=2!5 ! 5 !$
(iv) All the girls are never together $=10!-5 ! 6 !$
There are 10 professors and 20 lecturers, out of whom a committee of 2 professors and 3 lecturers is to be formed. Find
Column I | Column II | ||
---|---|---|---|
(i) | In how many ways committee can be formed? | (a) | ${ }^{10} C_2 \times{ }^{19} C_3$ |
(ii) | in how many ways a particular professor is included? | (b) | ${ }^{10} C_2 \times{ }^{19} C_2$ |
(iii) | in how many ways a particular lecturer is included? | (c) | ${ }^{9} C_1 \times{ }^{20} C_3$ |
(iv) | in how many ways a particular lecturer is excluded? | (d) | ${ }^{10} C_2 \times{ }^{20} C_3$ |
(i) We have to select 2 professors out of 10 and 3 lecturers out of $20={ }^{10} \mathrm{C}_2 \times{ }^{20} \mathrm{C}_3$
(ii) When a particular professor included $={ }^{10-1} C_1 \times{ }^{20} C_3={ }^9 C_1 \times{ }^{20} C_3$
(iii) When a particular lecturer included $={ }^{10} \mathrm{C}_2 \times{ }^{19} \mathrm{C}_2$
(iv) When a particular lecturer excluded $={ }^{10} \mathrm{C}_2 \times{ }^{19} \mathrm{C}_3$