Permutations and Combinations-algebra tutors

Permutations and Combinations(part-2)

In the previous post, we discussed the fundamental principle of counting and various methods of permutationsIn this post, I shall discuss combinations in details.

Meaning of Combination

If we are given a set of objects and we want to select a few objects out of this set, then we can do it by many different ways. These ways are known as combinations.
Example- If we are given three balls marked as B, W and R and we want to select two balls then we can select like this- BW, BR, WR.

These are known as the combination of this selection. This is an important concept in Online Maths Tutoring

Combination of n different objects taken r at a time when repetition is not allowed

If repetition is not allowed the number of ways of selecting r objects out of a group of n objects is called  {}^n{c_r}

{}^n{c_r}=   \frac{{n!}}{{r!\left( {n - r} \right)!}}

In latest notation system  {}^n{c_r} is also known as C(n;r) or   \left( \begin{array}{l} n\\ \\ r \end{array} \right)

Properties of  \left( \begin{array}{l} n\\ \\ r \end{array} \right)

It’s a very useful and interesting Mathematical tool. It has following properties.

(i) {}^n{c_r}{}^n{c_{n - r}}

(ii)  {}^n{c_n} = {}^n{c_0} = 1

(iii)  {}^n{c_r} + {}^n{c_{r - 1}} = {}^{n + 1}{c_r}    known as Pascal’s law

(iv) r. {}^n{c_r} = n{}^{n - 1}{c_{r - 1}}

(v)  \frac{{{}^n{c_r}}}{{{}^n{c_{r - 1}}}} = \frac{{n - r + 1}}{r}

(vi) If n is even then we should put r=n/2 for maximum value of  {}^n{c_r} and if n is odd then  {}^n{c_r} is greatest when r= \frac{{{n^2} - 1}}{4}

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(vii) In the expansions of {(1 + x)^n} if we put x=1 then

{}^n{c_0} + {}^n{c_1} + {}^n{c_2} + ....... + {}^n{c_n} = {2^n}

{}^n{c_0} + {}^n{c_2} + {}^n{c_4} + ......... = {2^{n - 1}}

{}^n{c_1} + {}^n{c_3} + {}^n{c_5} + ......... = {2^{n - 1}}

Some difficult Combinations

(a) If we want to select r objects out of a set of n objects such that p particular objects are always selected then the total number of ways =  {}^{n - p}{c_{r - p}}

(b) If we want to select r objects out of a set of n objects such that p particular objects are never selected then the total number of ways =  {}^{n - p}{c_r}

(c) If we want to select one or more object(at least one object) out of a set of n objects then the total number of ways= {2^n} - 1

(d) If we want to select zero or more object(at least zero object) out of a set of n objects then the total number of ways=  {2^n}

(e) If we want to select two or more object(at least two objects) out of a set of n objects then the total number of ways=  {2^n} - {}^n{c_1} - 1

A few more Combinations….

(f) If we want to select one or more object(at least one object) out of a set of n identical objects then the total number of ways= {}^n{c_1} = n

(g) If we have a set of objects in which p objects are of one type, q objects are of other type and r objects are of some other type and we want to select some or all out of this set then the total number of combinations=(p+1)(q+1)(r+1)-1

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(h) If we have a set of objects in which p objects are of one type, q objects are of other type, r objects are of some other type, n objects are different and we want to select at least one out of this set then total number of combinations=(p+1)(q+1)(r+1) {2^n}

(i) If we have a set of objects in which p objects are of one type, q objects are of other type, r objects are of other type, n objects are different and we want to select such that at least one object of each kind is included total number of combinations= pqr( {2^n} - 1)

(j) Number of ways in which (m+n+p) can be divided into three unequal groups=\frac{{(m + n + p)!}}{{m!n!p!}}

(h) Number of ways in which (m+n+p) can be divided and distributed into three unequal groups= \frac{{(m + n + p)!}}{{m!n!p!}}3!

Problems where we use both permutations and combinations

In exams, usually, we are asked questions in which we have to use both permutations and combinations.

(a) The number of ways of selecting and arranging r objects out of a group of n different objects such that p particular objects are always included= {}^{n - p}c_{r - p}^{}. r!

(b) The number of ways of selecting and arranging r objects out of a group of n different objects such that p particular objects are always excluded =  {}^{n - p}c_r^{}.r!

(c) The number of ways of selecting and arranging r objects out of a group of n different objects such that p particular objects are always seperated=  {}^{n - p + 1}{c_p}(n - p)!p!

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De-arrangement theorem

This theorem is used to permutate n distinct objects such that no object gets its original position.

according to this theorem, total no. of ways=  n!\left[ {1 - \frac{1}{{1!}} + \frac{1}{{2!}} - \frac{1}{{3!}}.....{{( - 1)}^n}\frac{1}{{n!}}} \right]

Formation of Geometrical Figures Using Combinations-

(a) If we are given n points on a plane then, the number of lines formed by joining these points =  {}^n{c_2}

(b) If we are given n points on a plane out of which m are in a straight line, the number of lines formed by joining these points= {}^n{c_2} - {}^m{c_2} + 1

(c) If we are given n points on a plane then, the number of triangles formed by joining these points =  {}^n{c_3}

(d) If we are given n points on a plane out of which m are in a straight line, the number of triangles formed by joining these points=  {}^n{c_3} - {}^m{c_3}

(e) If we are given n points on a plane, the number of diagonals formed by joining these points= {}^n{c_2} - n =  \frac{{n(n - 3)}}{2}

Here I am posting a pdf. Use concepts that are given in this post and the previous post of permutation and combination to solve questions given in the pdf

 Permutations and Combinations.pdf

<img src="demo.png" alt="demo">

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