# 61. Permutations and Combinations Solutions Part 4ΒΆ

We know that if \({}^nC_x ={}^nC_y\) then either \(x = y\) or \(x + y = n\)

Using this property in reverse manner \(^{20}C_{13} ={}^{20}C_7\) and \(^{20}C_{14} ={}^{20}C_6\)

Thus, \(^{20}C_{13} +{}^{20}C_{14} -{}^{20}C_6 -{}^{20}C_7 = 0\)

\(\frac{n!}{(n - r + 1)!(r - 1)!} = 36\)

\(\frac{n!}{(n-r)!r!} = 84\)

\(\frac{n!}{(n - r - 1)!(r + 1)!} = 126\)

From first two we have \(3n - 10r + 3 = 0\)

From last two we have \(2n - 5r - 3 = 0\)

Solving these two equations we get \(n = 9, r = 3\)

Total no. of ways of selecting \(4\) vertices out of \(n ={}^nC_4\)

Thus, \(^nC_4\) quadrilaterals will be formed.

Total no. of parties \(={}^7C_3 = 35\)

Total no. of ways he can invite them when a particular friend is always present \(= ^1C_1.{}^6C_2 = 15\)

Let us solve these one by one:

When a particular member is always selected then \(5\) selections out of remaining \(11\) can be done in \(^{11}C_5\) ways.

When a particular member is always excluded then \(6\) selections out of remaining \(11\) can be done in \(^{11}C_6\) ways.

Let us solve these one by one:

Total no. of ways of seating them \(={}^6P_6 = 6!\)

Treating C and D as one person so that they always sit together, total no. of ways of seating them \(={}^5P_5 = 5!\) but C and D can be arranged in \(2!\) ways among themselves making it \(5!2!\)

Thus, total no. of arrangements when C and D do not sit together \(= 6! - 5!2!\)

Since C is always included rest \(3\) positions can be selected from remaining \(5\) in \(^5C_3 = 10\) ways.

Since C is always included and E is always excluded \(3\) positions can be selected from remaining \(4\) in \(^4C_2 = 6\) ways.

Total no. of possible tickets \(={}^nC_2 = \frac{n(n - 1)}{2} = 105\)

\(\Rightarrow n^2 - n - 210 = 0\)

\(\Rightarrow (n - 15)(n + 14) = 0\)

But \(n\) cannot be negative \(\therefore n = 15\)

Total no. of ways of selecting \(2\) points from \(15 + 6\) \(={}^{21}C_2\)

Total no. of ways of selecting \(2\) points from \(6\) collinear points \(={}^6C_2\)

But these lines are actually only one line.

\(\therefore\) total no. of straight lines \(={}^{21}C_2 -{}^6C_2 + 1\)

Total no. of triangles formed \(={}^6C_3 +{}^6C_2.{}^{15}C_1 +{}^6C_1.{}^{15}C_2\)

**Case I:**When no points from collinear points are selected.Total no. of quadrilaterals formed \(={}^5C_4\)

**Case II:**When one point from collinear points are selected.Total no. of quadrilaterals formed \(={}^5C_3.{}^5C_1\)

**Case III:**When two points from collinear points are selected.Total no. of quadrilaterals formed \(={}^5C_2.{}^5C_2\)

Sum of these three cases gives us the total no. of quadrilaterals formed.

**Case I:**When two points from \(3\) interior points are selected.Total no. of triangles formed \(={}^3C_2.{}^4C_1 +{}^3C_2.{}^5C_1\)

**Case I:**When two points from \(4\) interior points are selected.Total no. of triangles formed \(={}^4C_2.{}^3C_1 +{}^4C_2.{}^5C_1\)

**Case I:**When two points from \(5\) interior points are selected.Total no. of triangles formed \(={}^5C_2.{}^4C_1 +{}^5C_2.{}^3C_1\)

Sum of these gives total no. of triangles formed.

No. of ways of selecting goalkeeper \(={}^2C_1\)

Rest \(10\) players can be selected out of remaining \(12\) in \(^{12}C_{10}\) ways.

\(2\) men can be chosen from \(5\) men in \(^5C_2\) ways.

\(2\) women can be chosen from \(6\) women in \(^6C_2\) ways.

Thus, total no. of selecting \(2\) men and \(2\) women from \(5\) men and \(6\) women \(={}^5C_2.{}^6C_2\)

Since all boys have to get at least one article at least one boy will receive two of them.

Thus, total no. of ways \(={}^8C_2.{}^6C_1.{}^5C_1.{}^4C_1.{}^3C_1.{}^2C_1.{}^1C_1\)

But the boy receiving two articles can be any of :math:`7`boys.

\(\therefore = 7({}^8C_2.{}^6C_1.{}^5C_1.{}^4C_1.{}^3C_1.{}^2C_1.{}^1C_1)\)

**Case I:**When \(3\) ladies are included.\(3\) ladies can be included in \(^4C_3\) ways.

Remaining \(2\) positions can be filled by \(7\) men in \(^7C_2\) ways.

**Case II:**When \(4\) ladies are included.\(4\) ladies can be included in \(^4C_4\) ways.

Remaining \(1\) position can be filled in \(^7C_1\) from \(7\) men.

Thus, total no. of ways \(={}^4C_3.{}^7C_2 +{}^4C_4.{}^7C_1\)

**Case I:**When the student answers \(3\) questions from each group.Total no. of ways \(={}^5C_3.{}^5C_3\)

**Case II:**When he answers \(4\) from first group and \(2\) from second.Total no. of ways \(={}^5C_4.{}^5C_2\)

**Case III:**When he answers \(2\) from first group and \(4\) from second.Total no. of ways \(={}^5C_4.{}^5C_2\)

Thus, total no. of ways \(= 2({}^5C_4.{}^5C_2) +{}^5C_3.{}^5C_3\)

Let us solve these one by one.

Since a particular professor is included, no. of ways of choosing \(1\) more professor out of remaining \(9 ={}^9C_1\)

No. of ways of picking \(3\) students out of \(20 ={}^{20}C_3\)

Total no. of ways of forming committee \(={}^9C_1.{}^{20}C_3\)

Since a particular professor is excluded, no. of ways of choosing

\(2\) professors out of remaing \(9 ={}^9C_2\)

No. of ways of picking \(3\) students out of \(20 ={}^{20}C_3\)

Total no. of ways of forming committee \(={}^9C_2.{}^{20}C_3\)

**Case I:**When only one girl is part of committee.No. of ways of choosing girls \(= ^7C_1\)

No. of ways of choosing boys \(= ^6C_4\)

Thus, no. of ways of forming the committee \(={}^7C_1.{}^6C_4\)

**Case II:**When two girls are part of committee.No. of ways of choosing girls \(={}^7C_2\)

No. of ways of choosing boys \(={}^6C_3\)

Thus, no. of ways of forming the committee \(={}^7C_2.{}^6C_3\)

Similarly, the committee may have three, four or five girls.

Total no. of ways of forming committee \(={}^7C_1.{}^6C_4 +{}^7C_2.{}^6C_3 + {}^7C_3.{}^6C_3 +{}^7C_4.{}^6C_1 +{}^7C_5{}.^6C_0\)

Let us solve these one by one:

Total no. of persons \(= 6 + 4 = 10\)

Since there is no restriction, total no. of ways of forming the committee \(= ^{10}C_5\)

Following 167 for this case, total no. of committees possible \(={}^6C_4.{}^4C_1 +{}^6C_3.{}^4C_2 +{}^6C_2.{}^4C_3 +{}^6C_1.{}^4C_4\)

Following like 167 total no. of ways of forming committee

\(={}^8C_1.{}^4C_4 +{}^8C_2.{}^4C_3 +{}^8C_3.{}^4C_2 +{}^8C_2.{}^4C_1\)

Following like 167. total no. of ways for selections

\(={}^{9}C_8.{}^6C_4 + {}^{9}C_7.{}^6C_5 + {}^9C_6.{}^6C_6\)

Problem no. 171 to 174 have been left as exercises.

Let us solve these one by one.

\(2\) points can be selected out of \(8\) and \(1\) from P and Q.

Total no. of such triangles \(= {}^8C_2.{}^2C_1\)

\(1\) point can be selected out of \(8\) and both P and Q.

Total no. of such triangles \(= {}^8C_1.{}^2C_2\)

When P is included and Q is excluded \(2\) points have to be chosen out of \(8\).

Total no. of such triangles \(= {}^8C_1.{}^2C_2\)

Total candidates are \(10\). Elector can vote for either \(2\) vacancies or \(1\) vacancy.

Thus, total no. of ways he can vote \(= {}^{10}C_2 + {}^{10}C_1 = 55\)

Problem no. 177 and 178 have been left as exercises.

From apples zero or more can be selected making it \(6\) choices.

From oranges zero or more can be selected making it \(5\) choices.

From mangoes zero or more can be selected making it \(4\) ways.

Total no. of choices \(= 120\) but one of these contain no fruits. Thus, desired answer \(= 119\)

Total no. of selections for red balls \(2^4 - 1\) as one red ball has to be always selected.

Total no. of selections for green balls \(2^3\)

Thus, desired answer \(= 15 . 8 = 120\)

There are three different coins and to make a sum at least one has to be selected. Thus, desired answer \(= 2^3 - 1 = 7\)

Following like previous problem, answer \(= 2^5 - 1 = 31\)

The voter can vote for \(1, 2\) or \(3\) candidates. Total no. of ways the voter can vote \(= {}^6C_3 + {}^6C_2 + {}^6C_1 = 41\)

Let there are vacancies for \(n\) members. So no. of candidates \(= n + 1\)

Total no. of ways of voting \({}^{n+1}C_n + {}^{n + 1}C_{n - 1}\ldots + {}^{n + 1}C_1\)

\(= 2^{n + 1} - 2 = 30 \Rightarrow n = 5\)

Following the formula for distribution, no. of ways \(= \frac{12!} {4!}^3\)

Rest of the problems are left as exercises.