Permutations and Combinations-algebra tutors

Permutations and Combinations(part-2)

In my previous post, we discussed the fundamental principle of counting and various methods of permutations. In 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.

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}

(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 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 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 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 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 total number of ways=  {2^n} - {}^n{c_1} - 1

(f) If we want to select one or more object(at least one object) out of a set of n identical objects then 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 total number of combinations=(p+1)(q+1)(r+1)-1

(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!

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 previous post to solve questions given in the pdf

 Permutations and Combinations.pdf

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