As we have seen earlier, if we have to pick out any two letters from the three letters a, b, c, we can do so in three different ways: ab, ac, bc. These are called combinations.

n, r are integers. If n ≥ r, for the set of n elements, a subset of r elements is called a combination.

Therefore, a combination is an unordered selection of r elements from a set of n elements.

The number of combinations of n dissimilar things taken r at a time is denoted by ^{n}C_{r} or n or C(n, r).

To find the number of combinations of “n” dissimilar things taken “r” at a time is

n!

^{n}C_{r} = ———

r! (n–r)!

Theorem

^{(n–1)}C_{(r–1)} + ^{(n–1)}C_{r} = ^{n}C_{r}

If m and n are positive integers with m ≠ n, then the total m + n distinct objects can be divided into two groups of m objects and n objects in

(m+n)!

——— ways

m! n!

(m + n + p) objects can be divided into three groups of m objects, n objects and p objects in

(m + n + p)!

————— ways.

m! n! p!

- If
^{n}C_{r}= 126 find n**Solution:**Since^{n}C_{r}is a positive integer, we have^{n}C^{r}= 126 = 63 ∗ 2 = 9∗7∗2

**∗**8**∗**7 9**∗**8**∗**7**∗**6

**∗**6

9**∗**8**∗**7**∗**6

= ——————

4**∗**3**∗**2**∗**1

^{9}C_{4}

∴n = 9 -
If
^{n}C_{6}=^{n}C_{8}, then what is^{n}C_{6}?

**Solution:**

Since^{n}C_{6}=^{n}C_{8}, we have n = 6 + 8 = 14

^{14}C_{9}=^{14}C_{6}

**∗**13**∗**12**∗**11**∗**10

**∗**4**∗**3**∗**2**∗**1 - In how many ways can a committee of 5 members be formed from 8 men and 5 women such that at least 2 women should be in each committee?

**Solution:**

If there are at least 2 women in each five-member committee, then the committees are- 3 men, 2 women
- 2 men, 3 women
- 1 man, 4 women
- No man, 5 women
- The number of selections related to =
^{8}C_{3}∗^{5}C_{2}= 56∗10 = 560 - The number of selections related to =
^{8}C_{2}∗^{5}C_{3}= 28∗10 = 280 - The number of selections related to =
^{8}C_{1}∗^{5}C_{4}= 8∗5 = 40 - The number of selections related to =
^{8}C_{0}∗^{5}C_{5}= 1∗1 = 1 - ∴ The number of ways of selecting a five-member committee is 560 + 280 + 40 + 1 = 881.

- There are 3 questions in the first section, 3 questions in the second section and 2 questions in third section on an exam. In how many ways can a student who appeared for the exam choose to answer any 5 questions of the paper, choosing at least 1 question from each section?

**Solution:**

Out of 8 questions, the student can select 5 questions in following ways,

The number of ways of selecting in 1st way

=^{3}C_{2}∗^{3}C_{1}∗^{2}C_{1}= 6

The number of ways of selecting in 2nd way

=^{3}C_{2}∗^{3}C_{2}∗^{2}C_{1}= 18

The number of ways of selecting in 3rd way

=^{3}C_{2}∗^{3}C_{1}∗^{2}C_{2}= 9

The number of ways of selecting in 4th way

=^{3}C_{1}∗^{3}C_{2}∗^{2}C_{2}= 9

The total number of ways of selection = 6 + 18 + 9 + 9 = 42. - Find the number of divisors of 21600.

**Solution**

The canonical form for 21600 is 21600 = 2^{5}∗3^{3}∗5^{2}

Among the divisors of the given number, the divisor 2 may not be there, or one 2 or two 2’s or three 2’s or four 2’s or five 2’s may be there. Therefore, five 2’s can be treated as 5 + 1 = 6 ways. Similarly, treat the divisor 3 in 4 ways, 5 in 3 ways. Hence, the required number of divisors = 6∗ 4∗3 = 72. Out of these, deleting 1 and 21600, the required number of divisors is 72 - 2 = 70.

- Find the following:

How many ways can 5 men and 3 women be seated in a row so that two women cannot sit side by side?

Answer:

Since there is no restriction on men, they can sit in a row in 5! = 120 ways

Since women cannot sit side by side they can sit in between two men or in the first place or in the last place.

X M X M X M X M X M X

If the sign ‘X’ shows the women place, there are 6 vacancies for women to sit,

In these 6 vacancies 3 women can sit in^{6}P_{3}= 120 ways

∴ The number of ways where 5 men and 3 women can sit with the

given condition is

120∗120 = 14400 -
Which of the following is the correct answer?
a. If
^{n}P_{6}= 30^{n}P_{4}find the value of n ?- n = 10
- n = 2
- n = 8

Then in how many ways all jobs can be filled?- 25
- 20
- 120

- 24
- 120
- 60

- 5040
- 7!
- 720

a. **Solution:**

a. n = 10

b. **Solution:**

c. 120

c. **Solution:**

a. 24

d. **Solution:**

c. 720

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