There will be only 24 selections containing atleast one red ball out of a bag containing 4 red and 5 black balls. It is being given that the balls of the same colour are identical.
Eighteen guests are to be seated, half on each side of a long table. Four particular guests desire to sit on one particular side and three others on other side of the table. The number of ways in which the seating arrangements can be made is $\frac{11!}{5!6!}(9!)(9!)$.
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 |
$\begin{aligned} & =\left(2^5-1\right) \times 2^7 \\ & =128 \times 31=3968\end{aligned}$