Question

10 example ⚡????of combination with solution

Answer 1

Combination is simply a manner of selection of members from a given group. In such kind of selection the order does not make a difference. Also, there is no repetition taken in the concept of combinations. They are assumed without repetition only. 

It is denoted by nCk,(Cnk)nCk,(Ckn) or most commonly C (n, k), where k is the numbers of members selected and n is the total number of members. Clearly, k ≤≤ n. If k>nk>n, then the combination, C(n,k)C(n,k) = 0. 

Combination problems are given below:
Example 1: In a lucky draw chits of ten names are out in a box out of which three are to be taken out. Find the number of ways in which those three names can be taken out.

Solution: The possible number of ways for finding three names out of ten from the box is:


C (10, 3) = 10!(3!7!)10!(3!7!) = 10∗9∗8∗7!7!∗3∗2∗110∗9∗8∗7!7!∗3∗2∗1= 72067206 = 120

So there are 120 different ways of choosing three names out of the ten from the box.

Example 2: Find the values of 14C5, 10C810C8 and C(7, 2).

Solution: We know that the formula for combination, i.e. C(n,k)C(n,k) = n!(k!(n−k)!)n!(k!(n−k)!)
14C5 = 14!(5!(14−5)!)14!(5!(14−5)!) = 14!(5! 9!)14!(5! 9!) = 14.13.12.11.10.9!(5.4.3.2.1.9!)14.13.12.11.10.9!(5.4.3.2.1.9!)  = 2002

 10C810C8 = 10!(8!(10−8)!)10!(8!(10−8)!) = 10!(8! 2!)10!(8! 2!) = 10.9.8!8!.2)10.9.8!8!.2)  = 45

C(7, 2) = 7!(2!(7−2)!)7!(2!(7−2)!) = 7!(5! 2!)7!(5! 2!) = 7.6.5!5!.2)7.6.5!5!.2) = 21

Example 3: Let us suppose we have 12 adults and 10 kids as an audience of a certain show. Find the number of ways the host can select three persons from the audiences to volunteer. The choice must contain two kids and one adult.

Solution: As order here does not matter so we have:

C (10, 2) * C (12, 1) = [10 * 9292] * [121121] = 45 * 12 = 540.

So there are 540 ways in which the host can choose the volunteers containing two kids and an adult.

Answer 2

Combination

Definition:

A combination is a method of picking elements from a collection in which the order in which they are chosen is irrelevant.

Formula:

${ }_{n} C_{r}=\frac{n !}{r !(n-r) !}$

${ }_{n} C_{r}=$ number of combinations,

$n=$ total number of objects in the set,

$r=$  number of choosing objects from the set.

where 0 ≤ r ≤ n.

Important result of combinations:

• The number of ways of selecting $n$objects out of $n$ objects is:

${ }^{n} C_{n}=\frac{n !}{n !(n-n) !}=\frac{n !}{n ! 0 !}=1$

• The number of ways of selecting $O$objects out of $n$objects is:

${ }^{n} C_{0}=\frac{n !}{0 !(n-0) !}=\frac{n !}{0 ! n !}=1$

• The number of ways of selecting 1 object out of $n$ objects is:

${ }^{n} C_{1}=\frac{n !}{1 !(n-1) !}=\frac{n \times(n-1) !}{(n-1) !}=n$

• ${ }^{n} C_{r}={ }^{n} C_{n-r}$

• ${ }^{n} C_{r}+{ }^{n} C_{r-1}={ }^{n+1} C_{r}$

Example:

Example 1: A pair must be formed from a group of five persons. The following formula can be used to determine the number of possible combinations.

${ }^{5} \mathrm{C}_{2}=\frac{5 !}{2 !(5-2) !}=\frac{5 !}{2 ! 3 !}=\frac{120}{2 \times 6}=10$

Example 2: There are a total of 4 letter combinations that may be constructed with the letters in the word DRIVEN.

${ }^{6} \mathrm{C}_{4}=\frac{6 !}{4 !(6-4) !}=\frac{6 !}{4 ! 2 !}=\frac{720}{24 \times 2}=15$

To know more about, example of permutation and combination, here

https://brainly.in/question/36535787?msp_poc_exp=2

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