Algebra Combinations

Algebra is a part of the mathematics in which, the infinite number of operations on equations, polynomials, inequalities, radicals, rational numbers, logarithms, etc. Combinations and also the permutations are the topics included in the algebra. As the name clearly indicates that the combinations states how many forms can a particular group  be combined irrespective of the order, while permutations is also the same but, it is concerned with the order. This article has the information about the algebra combinations in the mathematics.

 

Algebra Combinations:

 

Combinations:

            In general the combination is understandable, stating that how many ways a group of things or elements can be arranged irrespective of the order in which, they are combined. In math, the specific representation of the combinations is used and a specific formula to find the combinations possible from a group is derived. The formulae and the representation is as follows,

Representation for the Combinations =  `nCr`

where, n is the number of elements present totally.

           r is the number of elements picked for combination.

Formula  for finding the Combinations, `nCr` = ` (n!)/ (r!(n-r))!`

 Certain Identities of Combinations:

`nCn ` = 1,   `nC0` = 1

 

Example problems for algebra combinations:

 

Example 1 for algebra combinations:

Evaluate the algebra combination for  11C4  using the combinations formula.

Solution:

The formula used to calcualte teh  combinations is `nCr` = ` (n!)/ (r!(n-r))!`

                                                           11C4 = ` (11!)/ (4!(11-4))!`

                                                           11C4  ` (11!)/ (4!(7))!`

11C4 =  `(11*10*9*8) /( 4*3*2*1)`

11C4 = 330

 

Example 2  for algebra combinations:

Out  of 5 men 2 women, a committee of 3 is to be formed. In how many can it be formed if at lest one woman is to be included?

Solution:

The committee can be formed by choosing

(1) 1 woman and 2 men

(2) 2 women and 1 man

Now 1 woman out of 2 , and 2 men out of 5 may be choosen in

2C1 * 5C2 = (2(5*4)/(2*1)) 

                    = 20 ways

And 2 women out of 2, and 1 man out of 5 may chosen in they way,

2C2 * 5C1 = 1*5 =5 ways

 Total number of ways (20+5) = 25.

The number of ways in which atleast one  women is in the commitee is 25.

 

Example 3 for algebra combinations:

A committee of 5 is to formed out of 6 men and 4 ladies . In how many ways can this be done, when

(a) at least 2 ladies are included

Solution:

(a) We have to make a selection of,

(1) 2 ladies out of 4 and 3 men out of 6 or

(2) 3 ladies out of 4 and 2 men out of 6 or

(3) 4 ladies out of 4 and 1 men out of 6

The number of ways of these selections are

Case 1: 4C2 * 6C3 = 6 * 20 =120

Case 2: 4C3 * 6C2 = 4 * 15 = 60

Case 3: 4C4 * 6C1 = 1 * 6 = 6

The required number of ways  = 120+60+6 = 186.

The number of ways in which the two ladies included in the commitee is 186.