# 57. Permutations and Combinations Solutions 2¶

1. Ten thousands place can be filled in $$4$$ ways using $$1, 2, 3, 4$$ because if we put $$0$$ at that place then it will become a $$4$$ digit number.

Rest of the $$4$$ places can be filled in $$^4P_4$$ i.e. $$24$$ ways.

Thus total no. of $$5$$ digit numbers $$= 96$$

2. Total no. of such numbers $$= 7\times 6\times 5 = 210$$ as hundred’s place can be filled in $$7$$ ways, ten’s place can be filled in $$6$$ remaining ways and unit’s position can be filled in $$5$$ remaining ways.

3. Hundred’s place can be filled in $$5$$ ways with any digit occupying that place except $$0$$. Ten’s place can also be filled in $$5$$ ways with remaining $$5$$ digits and unit’s place can be filled in $$4$$ ways.

Thus, desired number $$= 5\times 5\times 4 = 100$$

4. Leftmost place can be filled in $$9$$ ways with the exception of $$0$$ not occupying that place. Rest of the places can be filled using $$9, 8, 7, 6, 5, 4, 3, 2$$ ways using rest of the remaining digits.

Thus, total no. of such numbers $$= 9\times 9!$$

5. Thousand’s place can be filled in $$5$$ ways with $$0$$ not occupying that place. Rest of the places can be filled in $$5, 4, 3$$ ways with remaining digits.

Thus, total no. of numbers $$= 5\times 5\times 4\times 3 = 300$$

6. Thousand’s place can be filled only by $$5$$ and $$9$$ making it $$2$$ ways to fill that place. Rest of the places can be filled in $$3, 2, 1$$ ways with remaining digits.

Thus, total no. of numbers $$= 2\times 3\times 2 = 12$$

7. Case I: When the number is of three digits.

Hundred’s place can be filled by $$3, 4, 5$$ in three ways. Rest two places can be filled in $$5$$ and $$4$$ ways with remaining digits.

Thus, total no. of three digit numbers $$= 3\times 5\times 4 = 60$$

Case II: When the number is of four digits.

Thousand’s place can be filled by $$1, 2, 3$$ in three ways. Rest of the three places can be filled in $$5, 4, 3$$ ways with remaining digits.

Thus, total no. of numbers $$= 3\times 5\times 4\times3 = 180$$

Therefore, desired answer $$= 60 + 180 = 240$$

8. For the number to be divisible by $$5$$ unit’s place has to either $$0$$ or $$5$$.

Case I: When $$0$$ occurs at units place.

Ten thousand’s place can be filled in $$5$$ ways, thousand’s place in $$4$$ ways and so on.

Thus total no. of such numbers $$= 5\times 4\times 3\times 2 = 120$$

Case II: When $$5$$ occurs at units place.

Ten thousand’s place can be filled in $$3$$ ways as $$0$$ cannot occupy that position. Rest of the place can be filled in $$3, 2, 1$$ ways with remaining digits.

Thus total no. of such numbers $$= 3\times 3\times 2 = 18$$

Desired answer $$= 120 + 18 = 138$$

9. Total no. of six digits number $$= 6! = 720$$

For the number to be not divisible by $$5$$ unit’s place must not be $$5$$. Thus, unit’s place can be filled in $$5$$ ways. Rest of the places can be filled in $$5!$$ or $$120$$ ways. Thus, no. of numbers not divisible by $$5 = 5\times5! = 600$$

10. For the number to be even unit’s place must be occupied by $$2$$ and $$4$$. Rest of the $$4$$ remaining places can be filled in $$4!$$ ways. Thus, total no. of such numbers $$= 4.4! = 96$$

11. Case I: When the number is of one digit.

Only $$5$$ is one positive integer divisible by $$5$$.

Case II: When the number is of two digits.

Unit’s place must be occupied by $$0$$ or $$5$$ making it possible to fill unit’s place in $$1$$ way each.

Ten’s place can be filled in $$8$$ ways or $$9$$ ways with remaining depending on what is at unit’s place.

Thus, total no. of numbers $$= 8 + 9 = 17$$

Case III: When the number is of three digits.

When $$0$$ occupies unit’s place, hundred’s and ten’s places can be filled in $$9$$ and $$8$$ ways respectively.

When $$5$$ occupies unit’s place, hundred’s and ten’s places can be filled in $$8$$ and $$8$$ ways respectively.

Thus, total no. of numbers $$= 9\times 8 + 8\times 8 = 136$$

Desired no. $$= 1 + 17 + 136 = 154$$

12. Hundred’s place can be filled in $$5$$ ways as $$0$$ cannot occupy that place. Then, ten’s place can be filled in $$5$$ ways and unit’s place can be filled in $$4$$ ways.

Total no. of numbers $$= 5\times 5\times 4 = 100$$

For the numbers to be odd unit’s place has to be filled using $$5$$ or $$7$$ making it $$2$$ ways.

Then, hundred’s position can be filled in $$4$$ ways and ten’s place in $$4$$ ways.

Thus total no. of odd numbers $$= 4\times 4\times 2 = 32$$

13. Case I: When the no. is of one digit.

Even numbers would be $$0, 2, 4$$.

Case II: When the no. is of two digits.

When unit’s place occupies $$0$$ ten’s place can be filled in $$4$$ ways. If unit’s place occupied $$2$$ or $$4$$ ten’s place can be filled in $$3$$ ways.

Thus total no. of numbers $$= 4 + 3 + 3$$

Similarly, rest of the problem can be solved and has been left as exercise.

14. If $$5$$ always occupies ten’s place then rest of the $$5$$ positions can be filled using remaining $$5$$ digits in $$5!$$ i.e. $$120$$ ways.

15. Let us solved both of sub-questions.

1. Number of four digit numbers $$= ^7P_4 = 840$$

2. Case I: When the thousand’s place is $$3$$.

Hundred’s place can be filled in $$4$$ ways using $$4, 5, 6, 7$$

Ten’s place can be filled in $$5$$ ways with $$5$$ remaining digits.

Unit’s place can be filled in $$4$$ ways.

Thus, total no. of numbers $$= 4\times 5\times 4 = 80$$

Case II: When the thousand’s place contains $$4, 5, 6, 7$$

Remaining places can be filled in $$6, 5, 4$$ ways with remaining digits.

Thus, total no. of numbers $$= 4\times 6\times 5\times 4$$

$$= 480$$

Thus, total no. of numbers greater than $$3400 = 80 + 480 = 560$$

16. Since thousand’s and unit’s place digits are fixed rest two positions can be filled in $$3$$ and $$2$$ ways with remaining digits.

Thus, total no. of such numbers $$= 3\times 2 = 6$$

Problem no. 67, 68, 69 and 70 are left as exercises.

1. Ten thousand’s place cannot be occupied by $$0$$.

When $$2$$ occurs at ten thousand’s place, no. of numbers $$^4P_4 = 24$$

When $$2$$ occurs at hundred’s, ten’s or unit’s place, no. of numbers $$= 3\times 3\times 2 = 18$$

$$\therefore$$ Sum of numbers $$= 24(2 + 4 + 6 + 8)\times 10000 + 18(2 + 4 + 6 + 8)\times 1000 + 18(2 + 4 + 6 + 8)\times 100$$ $$+ 18(2 + 4 + 6 + 8)\times 10 + 18(2 + 4 + 6 + 8)\times 1$$

$$= 5199960$$

Problem no. 72 and 73 have been left as exercises.

1. Each letter can be posted in $$4$$ ways. Thus, $$5$$ letters can be posted in $$4^5 = 1024$$ ways.

2. Each prize can be given in $$5$$ ways. Thus, $$3$$ prizes can be given in $$5^3 = 125$$ ways.

3. Each thing can be given in $$p$$ ways. Thus, $$n$$ things can be given in $$p^n$$ ways.

4. Each monkey can have $$m$$ masters. Thus, $$n$$ monkeys can have $$n^m$$ masters.

5. $$2$$ prized in Mathematics and Physics can be given in $$2^{10}$$ ways each. $$1$$ prize in Chemistry can be given in $$10$$ ways.

Thus, total no. of ways $$= 2^{10} + 2^{10} + 10 = 2058$$

6. $$12$$ cows can be loaded in $$12^{12}$$ ways.

$$12$$ calves can be loaded in $$12^{12}$$ ways.

$$12$$ horses can be loaded in $$12^{12}$$ ways.

Thus, total no. of ways $$= 3\times 12^{12}$$

7. Each delegate can be put in $$6$$ different ways.

Thus, $$5$$ delegates can be put in $$6^5$$ ways.

8. Ten thousand’s place can be filled in $$4$$ ways. Rest of the places can be filled in $$5$$ ways.

Thus, total no. of numbers $$= 4\times 5^4$$

9. Each ring can be had in $$4$$ fingers.

$$6$$ rings can be had in $$4^6$$ ways.

10. Thousand’s place can be filled using $$3, 4, 5$$ i.e. $$3$$ ways.

Rest of the places can be filled in $$6$$ ways each using any of the digits.

Thus, total no. of numbers $$= 3\times 6^3 = 648$$

11. Maximum no. of cars that can be numbered $$= 9^3 + 9^4$$

12. All question can be answered in $$4$$ ways. Thus, total no. of possible answers $$= 4^10$$ ways.

If no consecutive questions to be answered in same way then first question can be answered in $$4$$ ways while rest can be answered in $$3$$ ways. Thus, total no. of answers $$= 4\times 3^9$$

13. Treating all volumes as one book we have four books, which can be arranged in $$4!$$ ways. But books with three volumes can be arranged in $$3!$$ ways among themselves and books with two volumes can be arranged in $$2!$$ ways.

Thus, total no. of arrangements $$= 4!3!3!2!2!$$

14. Treating all copies as one book we have $$14$$ books which can be arranged in $$14!$$ ways. Since copies are identical they can be arranged in $$1$$ way among themselves.

Thus, total no. of arrangements $$= 14!$$

15. Treating persons of same nationality as one person we have three persons, which can be seated in $$3!$$ ways. But $$10$$ Indians can be seated in $$10!$$ ways, $$5$$ Americans and $$5$$ British can be seated in $$5!$$ ways.

Desired answer $$= 3!10!5!5!$$

16. Let us fix the positions of buys first.

xBxBxBxBxBxBx

B represents boys position while x are the positions that can be occupied by girls. $$6$$ boys can be seated in $$6!$$ ways. $$4$$ girls can be seated in $$^7P_4$$ ways.

Thus, desired answer $$=6\times ^7P_4$$

17. $$n$$ books can be arranged in $$n!$$ ways. Treating two particular books as one; when they occur together; total no.of arrangements $$=(n - 1)!$$ but $$2$$ books can be arranged in $$2$$ ways among themselves making it $$2(n - 1)!$$

Thus, desired answer $$= n! - 2(n - 1)! = (n - 2)(n- 1)!$$

18. Following previous example; setting $$n =\ 6$$; $$4.5! = 480$$

Rest of the problems are left as exercises.