# 66. Permutations and Combinations Solutions Part 3

1. The word “Independence” has $$12$$ letters with $$3$$ ns, $$2$$ ds, $$4$$ es and one each of i, p and c.

Thus, total no. of permutations possible $$= \frac{12!}{3!2!4!}$$

2. The word “Principal” has $$9$$ letters with:math:2 ps, $$2$$ is, and one each of r, n, c, a and l.

Taking all three vowels together as one total no. of permutation is

$$\frac{7!}{2!}$$ but the vowels themselves can be arranged in $$\frac{3!}{2!}$$ ways.

$$\therefore$$ desired answer $$= \frac{7!3!}{2!2!}$$

3. Mathematics has $$11$$ letters with m coming twice, a coming twice, t coming twice, h, e, t, i, c, s each coming once.

Therefore, total no. of words that can be formed using these $$= \frac{11!}{2!2!}$$

Treating vowels and consonants together no. of words that can be formed $$= 2!. \frac{8!3!}{2!2!2!}$$

That is because $$8$$ consonants can be arranged in $$\frac{8!}{2!2!}$$ ways.

$$3$$ vowels can be arranged in $$\frac{3!}{2!}$$ ways.

Finally, the consonants and vowels can be arranged in $$2!$$ ways when taken together.

4. The word “Director” has $$8$$ letters with only $$r$$ repeating for $$2$$ times. Taking three vowels as one we can arrange them in $$\frac{6!}{2!}$$ ways but three vowels can be arranged in $$3!$$ ways among themselves.

Thus, desired answer $$= \frac{3!6!}{2!}$$

5. The word “Plantain” has $$8$$ letters with a and n repeating $$2$$ times each. Treating all vowels as one word we have total no. of arrangements as $$\frac{6!}{2!}$$ but $$3$$ vowels can be arranged in $$\frac{3!}{2!}$$ ways among themselves.

Thus, desired answer $$= \frac{3!6!}{2!2!}$$

6. The word “Intermediate” has $$12$$ letters with i coming twice, t coming twice, e coming thrice, n, r, m, d, a coming once.

So we have $$6$$ consonants and $$6$$ vowels.

Sinec the relative order does not change consonants will occupy consonants place and vowels will occupy vowels place.

Therefore, total no. of desired words $$= \frac{6!6!}{2!2!3!}$$

7. The word “Parallel” has $$8$$ letters with l coming thrice, a coming twice, p, r, e each coming once.

Thus, total no. of arrangements $$= \frac{8!}{3!2!}$$

Treating all l as one; no. of numbers when all l come together $$= \frac{6!}{2!3!}$$

$$\therefore$$ Desired answer $$= \frac{8!}{3!2!} - \frac{6!}{2!31}$$

8. Let us solve for each case one by one.

1. In this case position of D is fixed. So, remaining $$4$$ letters can be arranged in $$4!$$ ways as there is no repetition.

2. Same as previous case $$4!$$ is the answer.

3. Same as previous case $$4!$$ is the answer.

4. In this case position of two letters is fixed so answer is $$3!$$

9. The word “violent” has $$7$$ letters. If vowels are to occupy all odd places then $$4$$ consonants can be arranged in $$4!$$ ways and vowels can be arranged in $$3!$$ ways.

Thus, desired answer $$= 4!3!$$

10. The word $$Saloon$$ has $$6$$ letters with o repeating twice.

Since, vowels and consonants occupy alternate places total no. of such words $$= 2\frac{3!3!}{2!}$$ as $$3$$ consonants can be arranged in $$3!$$ ways for $$3 places$$ and vowels can be arranged in $$\frac{3!}{2!}$$ ways. But then odd positions and even positions can be swapped doubling the no. of words formed.

11. The word $$Article$$ has $$7$$ letters with no repetitions.

Following as 109 total no. of such words $$= 4!3!$$

12. Since the number has to be greater than $$4$$ million the most significant place can be occupied by either $$4$$ or $$5$$

Case I: When most significant digit is $$4$$

Rest of the digits can arrnaged in $$\frac{6!}{2!2!}$$ ways as both $$3$$ and $$5$$ repeat twice.

Case II: When most significant digit is $$5$$

Total no. of numbers $$= \frac{6!}{2!}$$

Therefore, desired answer $$= \frac{6!}{2!2!} + \frac{6!}{2!}$$

13. Total no. of numbers $$= \frac{7!}{3!2!}$$ as $$2$$ repeats thrice and $$3$$ repeats twice.

For the no. of be odd unit’s place must be occupied by one of $$1, 3, 5$$.

Case I: When unit’s place is occupied by $$1$$ or $$5$$

Total no. of numbers $$= 2 \frac{6!}{3!2!}$$

Case II: When unit’s place is occupied by $$3$$

Total no. of numbers $$= \frac{6!}{3!}$$

Now total no. of odd numbers is sum of these two cases.

14. There are four odd places with $$4$$ odd digits with both $$1$$ and $$3$$ reapeting twice.

Also, there are three even places with $$2$$ repeating twice and $$4$$ coming only once.

Thus, total no. of numbers $$= \frac{4!}{2!2!}\frac{3!}{2!}$$

15. Case I: When $$1$$ occurs at most significant place and the number is of $$5$$ digits.

Remaining $$4$$ positions can be filled in $$^5P_4$$ ways.

Case II: When $$1$$ occurs at most significant place and the number is of $$6$$ digits.

Remaining $$5$$ positions can be filled in $$^5P_5$$ ways.

Case III: When one of $$2, 3, 4, 5$$ occurs at most significant place and the number is of $$5$$ digits.

Remaining $$4$$ positions can be filled in $$\frac{^5P_4}{2!}$$ ways.

Thus, total no. of such numbers $$= 4.\frac{^5P_4}{2!}$$

Case IV: When one of $$2, 3, 4, 5$$ occurs at most significant place and the number is of $$6$$ digits.

Remaining $$5$$ positions can be filled in $$\frac{5!}{2!}$$ ways.

Thus, total no. of such numbers $$= 4.\frac{5!}{2!}$$

Desired answer is sum of these.

16. Most significant place can be filled using one of $$1, 2, 3, 4, 5$$ i.e. in $$5$$ ways. Rest of the places can be filled in $$6$$ ways.

Thus, total no. of four digit numbers $$= 5\times 6^3$$

Total no. of four digit numbers without any digit repeating $$= 5\times 5\times 4\times 3$$

The difference of these two is the no. of numbers where at least one digit repeats.

17. Total no. of signals $$= \frac{9!}{5!2!2!}$$

18. Total no. of signals $$= ^6P_1 +~^6P_2 +~^6P_3 +~^6P_4 +~^6P_5 +~^6P_6$$

19. Total no. of arrangements $$= 5!$$. In half of these $$e$$ will come before $$i$$.

20. $$4!$$

21. Let us solve each case:

1. $$9!$$ because there is no restriction from the formula for circular permutation.

2. Let us first fix position of boys. They can be seated in $$4!$$ ways from the formula of circular permutation.

This will make $$5$$ empty seats between boys and girls can be seated in $$^5P_5$$ ways.

Therefore, total no. $$= 4!5!$$

22. Treating all girls as a single entity total no. of seating arrangements $$= 6!$$ from the formula of circular permutation as there will be $$7$$ entities to be seated.

But the $$4$$ girls can be seated in $$4!$$ ways among themselves.

Thus, total no. of numbers $$= 6!4!$$

23. If the line starts with a girl then total no. of ways $$= 5!5!$$

If the line starts with a boy then total no. of ways $$= 5!5!$$

Thus, total no. of ways $$= 2.5!5!$$

Second Part: First let us make the boys take up places. So $$5$$ boys can be arranged in $$4!$$ ways. Now $$5$$ girls can be seated in $$5$$ empty seats in $$5!$$ ways.

Thus, total no. of seating arrangements $$= 4!5!$$

24. Following as last problem, total no. of ways $$= 5!\times~^6P_5$$

25. Since clockwise or anti-clockwise arrangement does not matter in a necklace total no. of ways $$= \frac{49!}{2}$$

26. Treating two particular delegated and host as one person we have $$19$$ persons.

Total no. of ways of seating $$19$$ persons around a round table $$= 18!$$

But the delegated and host can be arranged in $$2!$$ ways as two delegated sit on either side of host.

Thus, total no. of desired arrangements $$= 18!2!$$

27. Four gentlemen can be seated in $$3!$$ ways. This will make $$4$$ empty seats between gentlemen so that no two ladies sit together.

No. of ways of making ladies sit in these $$4$$ positions $$= 4!$$

Thus, total no. of arrangements $$= 3!4!$$

28. Following as previous problem; total no. of arrangements $$= 6!~^7P_6$$

29. We know that if $$^nC_x =~^nC_y$$ then either $$x = y$$ or $$x + y = n$$

$$\therefore$$ Either $$3r = r+3\Rightarrow r = \frac{3}{2}$$ which is not possible as $$r$$ is an integer.

$$\therefore 3r + r + 3 = 15 \Rightarrow r = 3$$

30. $$^nC_6:^{n - 3}C_3 = 33:4$$

$$\therefore \frac{n!}{6!(n-6)!}\times\frac{3!(n - 3 - 3)}{(n-3)!} = \frac{33}{4}$$

$$\therefore \frac{n!}{(n - 3!)}.\frac{3!}{6!} = \frac{33}{4}$$

$$n(n - 1)(n - 2) = 6.5.33 = 11.10.9$$

$$n = 11$$

31. Given expression is $$^{47}C_4 + (^{51}C_3 + ^{50}C_3 + ^{49}C_3 + ^{48}C_3 + ^{47}C_3)$$

$$= ^{48C_4} + (^{51}C_3 + ^{50}C_3 + ^{49}C_3 + ^{48}C_3) [\because ^{47C_4 + ^{47}C_3} = ^{48}C_4]$$

Following similarly, $$= ^{52}C_4$$

32. Let $$p$$ be the product of $$r$$ consecutive integers beginning with $$n$$

Then $$p = n(n + 1)(n + 2) \ldots (n + r - 1)$$

$$\frac{p}{r!} = \frac{n(n + 1)(n + 2) \ldots (n + r - 1)}{r!}$$

$$= ^{n + r - 1}C_r$$

$$=$$ things taken $$r$$ at a time = an integer

Therefore, product is divisible by $$r!$$

33. Total no. of triangles formed $$= ^mC_3$$

34. Total no. of ways of choosing $$3$$ children out of $$8 = ^8C_3 = 56$$

Therefore, the man will have to go to zoo for $$56$$ times.

Second Part: Choosing $$3$$ children out of $$8$$ when a particular child is always chosen $$= ^1C_1.^7C_2 = 21$$

35. Choosing a pair of students out of $$n = ^nC_2$$

Total no. of cards sent by one pair $$= 2$$

Therefore, total no. of cards send $$= 2.^nC_2 = 600$$

$$n^2 - n - 600 = 0$$

$$n = -24, 25$$

But $$n$$ cannot be negative. Thus, $$n = 25$$

36. Whenever two vertices are joined either a diagonal or a side is formed.

Total no. of diagonals and sides $$= ^mC_2$$

But out of these $$m$$ will be sides.

Therefore, no. of diagonals $$= ^mC_2 - m = \frac{m(m - 1)}{2} - m$$

$$= \frac{m(m - 3)}{2}$$

37. Total no. of persons $$= 10$$

Choosing $$5$$ persons out of $$10 = ^{10}C_5$$

Choosing $$5$$ persons out of $$6$$ men $$= ^6C_5$$

Thus, no. of committees where at least one women is selected $$= ^10C_5 - ^6C_5$$

38. Let us solve these one by one:

1. Total no. of selections of $$3$$ points out of $$10 = ^{10}C_3$$

Total no. of selection of $$4$$ points out of $$4$$ collinear points $$= ^4C_3$$

Thus, no. of triangles formed $$= ^{10}C_3 - ^4C_3$$

2. Total no. of selections of $$2$$ points out of $$10$$ points $$= ^{10}C_2 = 45$$

Number of selections of two points when only one straight line is formed $$= ^4C_2 = 6$$

i.e. the two points are selected from $$4$$ collinear points making the straight line only one.

Therefore, total no. of straight lines $$= 45 - 6 + 1 = 40$$

3. Total no. of selections of $$4$$ points out of $$10 = ^{10}C_4 = 210$$

Number of selections when no quadrilateral is formed $$= ^4C_3.^6C_1 + ^4C_4.6C_0 = 25$$

Thus, total no. of quadrilaterals formed $$= 210 - 25 = 185$$

39. Case I: When fruits of same type are different.

Number of selections $$= (^4C_0 + ^4C_1 + \ldots + ^4C_4)$$ $$(^5C_0 + ^5C_1 + \ldots + ^5C_5)(^6C_0 + ^6C_1 + \ldots + ^6C_6) - 1$$

$$= 2^4.2^5.2^6 - 1 = 2^15 - 1$$ Case I: When fruits of same type are same.

Zero or more oranges can be selected in $$5$$ ways.

Zero or more apples can be selected in $$6$$ ways.

Zero or more mangoes can be selected in $$7$$ ways.

Thus, total no. of ways $$= 5.6.7 = 210$$

But in one of these ways no fruit is selected, making answer $$209$$.

40. Following case I of previous problem we have total no. of ways $$= (2^5 - 1).(2^4 - 1).2^3 - 1 = 3720$$

41. Factors of $$21,600$$ are $$2, 2, 2, 2, 2, 3, 3, 3, 5, 5$$

Thus, total no. of divisors $$= 6.4.3 - 1 = 71$$

Total no. of divisors including $$1 = 72$$

42. No. of ways $$= ^5C_1 + ^5C_2 + ^5C_3 + ^5C_4 + ^5C_5 = 2^5 - 1 = 31$$

43. Required number $$= ^{12}C_4.^8C_4.^4C_4$$

44. Total no. of letters in the word “EXAMINATION” $$= 11$$ out of which A, N, I occur twice. Different letters are E, X, A, M, I, N, T, O which are eight in count.

Case I: When all are different.

Total no. of words formed $$= ^8C_4.4!$$

Case II When two are identical and two are different.

There are three possibilities of identical letters with A, N, I each occurring twice.

Total no. of words formed $$= 3.^7C_2.\frac{4!}{2!} = 756$$

Case III: When letters are pair of identical letters.

Total no. of words formed $$= 3.\frac{4!}{2!2!} = 18$$

45. $$\because 30 \neq 4; n = 30 + 4 = 34$$

46. $$\because 12 \neq 8; n = 12 + 8 = 20$$

$$^{20}C_{17} = \frac{20.19.18}{3.2} = 1140$$

$$^{22}C_{20} = \frac{22.21}{2} = 231$$

47. $$\because r \neq r + 2; r + r + 2 = 18 \Rightarrow r = 8$$

$$^8C_6 = 28$$

48. $$n(n - 1)(n - 2)(n - 3) = 15.4.3.2$$

$$n(n - 1)(n - 2)(n - 3) = 6.5.4.3 \Rightarrow n = 6$$

49. $$\frac{15!}{r!(15-r)!}.\frac{(r-1)!(16 - r)!}{15!} = 11:5$$

$$\Rightarrow r = 5$$

50. $$r = 5$$