# 61. Permutations and Combinations Solutions Part 4¶

1. 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, $$^{20}C_{13} +{}^{20}C_{14} -{}^{20}C_6 -{}^{20}C_7 = 0$$

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

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

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

From first two we have $$3n - 10r + 3 = 0$$

From last two we have $$2n - 5r - 3 = 0$$

Solving these two equations we get $$n = 9, r = 3$$

3. Total no. of ways of selecting $$4$$ vertices out of $$n ={}^nC_4$$

Thus, $$^nC_4$$ quadrilaterals will be formed.

4. Total no. of parties $$={}^7C_3 = 35$$

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

5. Let us solve these one by one:

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

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

6. Let us solve these one by one:

1. 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 $$= 6! - 5!2!$$

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

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

7. 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 $$\therefore n = 15$$

8. 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 $$={}^6C_3 +{}^6C_2.{}^{15}C_1 +{}^6C_1.{}^{15}C_2$$

9. 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.

10. 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.

11. No. of ways of selecting goalkeeper $$={}^2C_1$$

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

12. $$2$$ 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 $$={}^5C_2.{}^6C_2$$

13. 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:7boys.

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

14. 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 $$={}^4C_3.{}^7C_2 +{}^4C_4.{}^7C_1$$

15. 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 $$= 2({}^5C_4.{}^5C_2) +{}^5C_3.{}^5C_3$$

16. Let us solve these one by one.

1. 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 $$={}^9C_1.{}^{20}C_3$$

2. 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 $$={}^9C_2.{}^{20}C_3$$

17. 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 $$={}^7C_1.{}^6C_4 +{}^7C_2.{}^6C_3 + {}^7C_3.{}^6C_3 +{}^7C_4.{}^6C_1 +{}^7C_5{}.^6C_0$$

18. Let us solve these one by one:

1. Total no. of persons $$= 6 + 4 = 10$$

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

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

19. Following like 167 total no. of ways of forming committee

$$={}^8C_1.{}^4C_4 +{}^8C_2.{}^4C_3 +{}^8C_3.{}^4C_2 +{}^8C_2.{}^4C_1$$

20. 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.

1. Let us solve these one by one.

1. $$2$$ 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 $$= {}^8C_1.{}^2C_2$$

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

Total no. of such triangles $$= {}^8C_1.{}^2C_2$$

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.

1. 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 $$= 119$$

2. 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 $$= 15 . 8 = 120$$

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

4. Following like previous problem, answer $$= 2^5 - 1 = 31$$

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

6. Let there are vacancies for $$n$$ members. So no. of candidates $$= n + 1$$

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

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

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

Rest of the problems are left as exercises.