Permutations and Combinations

Clearly, number of words start with $A = {{4!} \over {2!}} = 12$

Number of words start with $L = 4! = 24$

Number of words start with $M = {{4!} \over {2!}} = 12$

Number of words start with $SA = {{3!} \over {2!}} = 3$

Number of words start with $SL = 3! = 6$

Note that, next word will be "SMALL"

Hence, the position of word "SMALL" is 58th.

For a four digit number the first place can be filled in 3 ways with 6 or 7 or 8 and the remaining four places in 4! ways i.e., 3 $\times$ 4! = 72.

For a five digit number it can be arranged in 5! ways,

$\therefore$ total number of integers = (72 + 120) = 192.

Given, six cards and six envelops are numbered $1,2,3,4,5$ and 6.

The number of arrangements when all six

cards are placed in wronge envelops

$\begin{aligned} & =\left| \!{\underline {\, {6} \,}} \right.\left(1-\frac{1}{\left| \!{\underline {\, {1} \,}} \right.}+\frac{1}{\left| \!{\underline {\, {2} \,}} \right.}-\frac{1}{\left| \!{\underline {\, {3} \,}} \right.}+\frac{1}{\left| \!{\underline {\, {4} \,}} \right.}-\frac{1}{\left| \!{\underline {\, {5} \,}} \right.}+\frac{1}{\left| \!{\underline {\, {6} \,}} \right.}\right) \\ & =\left| \!{\underline {\, {6} \,}} \right.\left[1-\frac{1}{1}+\frac{3}{\left| \!{\underline {\, {3} \,}} \right.}-\frac{1}{\left| \!{\underline {\, {3} \,}} \right.}+\frac{5}{\left| \!{\underline {\, {5} \,}} \right.}-\frac{1}{5}+\frac{1}{\left| \!{\underline {\, {6} \,}} \right.}\right] \\ & =\left| \!{\underline {\, {6} \,}} \right.\left[\frac{2}{\left| \!{\underline {\, {3} \,}} \right.}+\frac{4}{\left| \!{\underline {\, {5} \,}} \right.}+\frac{1}{\left| \!{\underline {\, {6} \,}} \right.}\right] \\ & =2.6 .5 .4+4.6+1 \\ & =240+24+1 \\ & =265 \\ \end{aligned}$

In the above dearrangement, there are 5 ways in which card number 1 is going wrong envelope i.e, other than envelope number 1. So, when card number 1 going in envelop number 2 is $\frac{265}{5}=53$ ways.

Hint:

In the total dearrangement of 6 cards there are 5 ways in which card number 1 is going other than envelope number 1.

For alike n objects, number of ways we can select zero or more objects = n + 1 and number of ways we can select at least one object = n

Given 10 identical white balls, 9 identical green balls and 7 identical black balls.

To find number of ways for selecting atleast one ball.

Number of ways to choose zero or more white balls = (10 + 1) [since, all white balls are mutually identical]

Number of ways to choose zero or more green balls = (9 + 1) [since, all green balls are mutually identical]

Number of ways to choose zero or more black balls = (7 + 1) [since, all black balls are mutually identical]

Hence, number of ways to choose zero or more balls of any colour = (10 + 1) (9 + 1) (7 + 1)

Also, number of ways to choose a total of zero balls = 1

Hence, the number, if ways to choose at least one ball (irrespective of any colour) = (10 + 1) (9 + 1) (7 + 1) $-$ 1 = 879

Given, $b_n$ denotes the number of $n$-digit integer formed by the digits 0, 1 or both such that $n$-digit integer ending with 1 and no consecutive digits are '0'.

$\therefore \quad b_6=$ six digit number ending with 1.

Like 1 ........... 1, and rest four places are filled as Case No. (I) : Use four ' 1 '

$\text { Case No. (I) : Use four ' } 1 \text { ' }$

$\underline 1 \underbrace {\underline 1 \,\underline 1 \,\underline 1 \,\underline 1 }_{}\underline 1 $

Number of ways = 1

Case No. (II) : Use three '1' and one '0'

$\underline 1 \underbrace {\underline 1 \,\underline 1 \,\underline 1 \,\underline 0 }_{}\underline 1 $

Number of ways $=\frac{4!}{3!}=4$

Case No. (III) : Use two '1' and two '0'

$\underline 1 \underbrace {\underline 1 \,\underline 0 \,\underline 1 \,\underline 0 }_{}\underline 1 $

or

$\underline 1 \underbrace {\underline 1 \,\underline 1 \,\underline 0 \,\underline 1 }_{}\underline 1 $

or

$\underline 1 \underbrace {\underline 0 \,\underline 1 \,\underline 1 \,\underline 0 }_{}\underline 1 $

No. of ways = 3

Hence, $b_6=1+4+3=8$

For $a_n$

Case I : If the unit digit is 1, and rest $(n-1)$ places are filled as $a_{n-1}$

$\mathrm{\underbrace {1\underline {} \,\underline {} \,\underline {} \,\underline {...} \,\underline {...} \,\underline {} }_{(n - 1)\,places}\underline 1}$

Case II : If the unit digit is 0, then tenth place must be 1 and rest $(n-2)$ places are filled as $a_{n-2}$

$\mathrm{\underbrace {1\underline {} \,\underline {} \,\underline {} \,\underline {} \,....\,\underline {} }_{(n - 2)\,place}\underline 1 \underline 0}$

$\begin{aligned} & \text { Hence, } a_n =a_{n-1}+a_{n-2} \\ \Rightarrow \quad & a_{17} =a_{16}+a_{15} \end{aligned}$

Here, 5 distinct balls are to be distributed amongst 3 persons so that each gets at least one ball. So, two possible cases arises

Case I : Two of the persons get one-one ball each and the third person gets three balls.

i.e.

Now, A can get the ball in ${ }^5 \mathrm{C}_1$ ways. After that, $B$ can get one ball in ${ }^4 C_1$ ways and then after $C$ can get three balls in ${ }^3 \mathrm{C}_3$ ways.

Hence, Total number of ways

$\begin{aligned} & ={ }^5 \mathrm{C}_1 \cdot{ }^4 \mathrm{C}_1 \cdot{ }^3 \mathrm{C}_3 \times \frac{3!}{2!} \\ & =5 \times 4 \times 1 \times \frac{3 \times 2 \times 1}{2 \times 1} \\ & =60 \end{aligned}$

Case II : Two of the persons get two-two balls each and the third person gets one ball.

i.e.

Now, A can get two balls in ${ }^5 \mathrm{C}_2$ ways. After that, $B$ can get 2 ball in ${ }^3 \mathrm{C}_2$ ways and then after $C$ can get 1 ball in ${ }^1 C_1$ way. Hence, total number of ways

$\begin{aligned} & ={ }^5 C_2 \cdot{ }^3 C_2{ }^1 C_1 \cdot \frac{3!}{2!} \\ & =\frac{5 \times 4}{2 \times 1} \times \frac{3 \times 2}{2 \times 1} \times 1 \times \frac{3 \times 2 \times 1}{2 \times 1} \\ & =90 \end{aligned}$

Hence, total number of ways to distribute 5 balls $=60+90=150$

Combination with Repetition

(i) Only two possible cases arises:

Case I :

Case II:

(iii) Use the concept of combination to find individual ways and then add up the total ways in each case.

Let XA, X_B, X_C and X_D represent number of balls present in box A, B, C and D respectively.

As no box can be empty so,

X_A $\ge$ 1, X_B $\ge$ 1, X_C $\ge$ 1 and X_D $\ge$ 1

$\Rightarrow$ X_A $-$ 1 $\ge$ 0, $\Rightarrow$ X_B $-$ 1 $\ge$ 0, $\Rightarrow$ X_C $-$ 1 $\ge$ 0 and $\Rightarrow$ X_D $-$ 1 $\ge$ 0

t_A $\ge$ 0, t_B $\ge$ 0, t_C $\ge$ 0 and t_D $\ge$ 0

According to the question,

X_A + X_B + X_C + X_D = 10

$\Rightarrow$ (X_A $-$ 1) + (X_B $-$ 1) + (X_C $-$ 1) + (X_D $-$ 1) = 6

$\Rightarrow$ t_A + t_B + t_C + t_D = 6

Now question becomes, box A, B, C, and D can have none or one or more balls and total balls are 6

From formula we know, n things can be distributed among r people in ${}^{n + r - 1}{C_{r - 1}}$ ways where each people can have either 0 or more things.

$\therefore$ 6 balls can be distributed among 4 boxes in ${}^{6 + 4 - 1}{C_{4 - 1}} = {}^9{C_3}$ ways where each box can have either 0 or more balls.

Therefore, Statement 1 is correct. The number of ways of choosing any 3 places from 9 different places is ${}^9{C_3}$ ways. But Statement - 2 is not the correct explanation of Statement - 1.

We need 3 points to create a triangle. With 10 points number of triangle possible ${}^{10}{C_3}$

Here 6 points are on the same line so we can't make any triangle with those 6 points.

So subtract ${}^{6}{C_3}$.

$\therefore$ $N = {}^{10}{C_3} - {}^6{C_3}$

$ = {{10\,.\,9\,.\,8} \over {1\,.\,2\,.\,3}} - {{6\,.\,5\,.\,4} \over {1\,.\,2\,.\,3}}$

$ = 120 - 20$

$ = 100$

Thus number of ways $ = ({}^3{C_2}) \times ({}^9{C_2}) = 3 \times {{9 \times 8} \over 2} = 108$

The two possible cases are as follows:

Case 1 : There are five 1's; one 2; one 3. Therefore, the number of numbers is 7!/5! = 42.

Case 2 : There are four 1's; three 2's. Therefore, the number of numbers is 7!/4! 3! = 35. Hence, the total number of numbers is 42 + 35 = 77

This problem is solved using gap method. As here no 'S' is adjacent to each other so we have to put them in the gap. So first write all the letters other than 'S' such a way that there is a gap between two letters.

Given word is MISSISSIPPI.

Here, I = 4 times, S = 4 times, P = 2 times, M = 1 time

_M_I_I_I_I_P_P_

Those seven letters M, I, I, I, I, P, P can be arranged in ${{7!} \over {4!2!}}$ ways

Those seven letters creates 8 gaps and we have to choose 4 gaps from those 8 gaps to put those four 'S' letters.

This can be done ${}^8{C_4}$ ways.

After placing those four 'S' letters we can arrange them in ${{4!} \over {4!}}$ ways.

Therefore, required number of words

$ = {{7!} \over {4!2!}} \times {}^8{C_4} \times {{4!} \over {4!}}$

$ = {{7\,.\,6!} \over {4!4!}} \times {}^8{C_4}$

$ = 7\,.\,{}^6{C_4}\,.\,{}^8{C_4}$

(A) Considering ENDEA as one group, remaining letters are N, O, E, L.

So, no. of permutations = 5!

(A) - (i)

(B) E occurs in 1^st and last positions. The remaining letters are N, N, D, A, O, E, L.

No. of permutation

$ = {{7!} \over {2!}} = {{7 \times 6} \over 2} \times 5! = 21! \times 5!$

(B) - (iv)

(C) D, L, N should not occur in last five positions

$\Rightarrow$ D, L, N should occur in 1^st four positions, but we have D, L, N, N.

So, ways of arranging D, L, N, N in 1^st four positions

$ = {{4!} \over {2!}} = 12$

Ways of arranging remaining E, E, A, O, E in last five positions $ = {{5!} \over {3!}} = 20$,

Total $=12\times20=240=2\times5!$

(C) - (iv)

(D) A, E, O occur in odd positions.

No. of odd positions = 5 and letters are E, E, E, A, O. i.e. 5

Ways of arranging those 5 letters in 5 odd positions $ = {{5!} \over {3!}} = 20$

Remaining 4 letters D, L, N, N can be arranged in remaining 4 positions in $ = {{4!} \over {2!}} = 12$ ways

Total no. of permutations $=20\times12=240=2\times5!$

(D) - (iii)

The total number of ways is

${}^{12}{C_4} \times {}^{12 - 4}{C_4} \times {}^{12 - 4 - 4}{C_4} = {}^{12}{C_4} \times {}^8{C_4} \times {}^4{C_4} = {{12!} \over {{{(4!)}^3}}}$

The letter of word COCHIN in alphabetic order are C, C, H, I, N, O

Fixing $1^{\text {st }}$ letter $\mathrm{C}$ and keeping $\mathrm{C}$ at second place, rest 4 can be arranged in 4 ! ways.

Similarly, the words starting with $\mathrm{CH}, \mathrm{CI}, \mathrm{CN}$ are 4 ! in each case

Then fixing first two letters as CO next four places, when filled in alphabetic order give the word COCHIN.

$\therefore$ Number of words coming before COCHIN are $4 \times 4$!

$=4\times24=96$