Permutations and Combinations

To find the position of the word QUEEN:

$\bullet$ The number of words starting with E is 4! = 24.

$\bullet$ The number of words starting with N is ${{4!} \over 2} = 12$.

$\bullet$ The number of words starting with QE is 3! = 6.

$\bullet$ Number of words starting with QN is ${{3!} \over 2} = 3$.

Therefore, the position of the word QUEEN is next to the sum, 24 + 12 + 6 + 3 = 45.

That is, the word 'QUEEN' will be on 46th position.

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.

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

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$

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}$

The total number of ways is

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