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 $={ }^{10} \mathrm{C}_2 \times{ }^{19} \mathrm{C}_3$
Using the digits $1,2,3,4,5,6,7$, a number of 4 different digits is formed. Find
| Column I | Column II | ||
|---|---|---|---|
| (i) | how many numbers are formed? | (a) | 840 |
| (ii) | how many numbers are exactly divisible by 2? | (b) | 200 |
| (iii) | how many numbers are exactly divisible by 25? | (c) | 360 |
| (iv) | how many of these are exactly divisible by 4? | (d) | 40 |
(i) Total numbers of 4 digit formed with digits 1, 2, 3, 4, 5, 6, 7
$$=7 \times 6 \times 5 \times 4=840$$
(ii) When a number is divisible by 2 . At its unit place only even numbers occurs.
$$\text { Total numbers }=4 \times 5 \times 6 \times 3=360$$
(iii) Total numbers which are divisible by $25=40$
(iv) A number is divisible by 4 , If its last two digit is divisible by 4 .
$\therefore$ Total such numbers $=200$
How many words (with or without dictionary meaning) can be made from the letters of the word MONDAY, assuming that no letter is repeated, if
| Column I | Column II | ||
|---|---|---|---|
| (i) | 4 letters are used at a time | (a) | 720 |
| (ii) | All letters are used at a time | (b) | 240 |
| (iii) | All letters are used but the first is a vowel. | (c) | 360 |
(i) 4 letters are used at a time $={ }^6 P_4=\frac{6!}{2!}=6 \times 5 \times 4 \times =360$
(ii) All letters used at a time $=6$ ! $=6 \times 5 \times 4 \times 3 \times 2 \times 1=720$
(iii) All letters used but first is vowel $=2 \times 5$ ! $=2 \times 5 \times 4 \times 3 \times 2 \times 1=240$