# 57. Permutations and Combinations Solutions 2¶

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 \(= 96\)

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.

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 \(= 5\times 5\times 4 = 100\)

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 \(= 9\times 9!\)

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 \(= 5\times 5\times 4\times 3 = 300\)

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 \(= 2\times 3\times 2 = 12\)

**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 \(= 60 + 180 = 240\)

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 \(= 120 + 18 = 138\)

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 \(5 = 5\times5! = 600\)

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 \(= 4.4! = 96\)

**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. \(= 1 + 17 + 136 = 154\)

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 \(= 4\times 4\times 2 = 32\)

**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.

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.

Let us solved both of sub-questions.

Number of four digit numbers \(= ^7P_4 = 840\)

**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 \(3400 = 80 + 480 = 560\)

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.

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.

Each letter can be posted in \(4\) ways. Thus, \(5\) letters can be posted in \(4^5 = 1024\) ways.

Each prize can be given in \(5\) ways. Thus, \(3\) prizes can be given in \(5^3 = 125\) ways.

Each thing can be given in \(p\) ways. Thus, \(n\) things can be given in \(p^n\) ways.

Each monkey can have \(m\) masters. Thus, \(n\) monkeys can have \(n^m\) masters.

\(2\) 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 \(= 2^{10} + 2^{10} + 10 = 2058\)

\(12\) 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 \(= 3\times 12^{12}\)

Each delegate can be put in \(6\) different ways.

Thus, \(5\) delegates can be put in \(6^5\) ways.

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 \(= 4\times 5^4\)

Each ring can be had in \(4\) fingers.

\(6\) rings can be had in \(4^6\) ways.

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 \(= 3\times 6^3 = 648\)

Maximum no. of cars that can be numbered \(= 9^3 + 9^4\)

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 \(= 4\times 3^9\)

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 \(= 4!3!3!2!2!\)

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 \(= 14!\)

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 \(= 3!10!5!5!\)

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 \(=6\times ^7P_4\)

\(n\) 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 \(= n! - 2(n - 1)! = (n - 2)(n- 1)!\)

Following previous example; setting \(n =\ 6\); \(4.5! = 480\)

Rest of the problems are left as exercises.