Permutation and Combination MCQ Questions & Answers in Algebra | Maths

The number will have 2 pairs and 2 different digits.
The number of selections $$ = {\,^4}{C_2} \times {\,^2}{C_2},$$   and for each selection, number of arrangements $$ = \frac{{6!}}{{2!\,2!}}.$$  Therefore, the required number of numbers $$ = {\,^4}{C_2} \times {\,^2}{C_2} \times \frac{{6!}}{{2!\,2!}}.$$

Required number of ways
$$\eqalign{ & \sum\limits_{r = 2}^5 {^5{C_{5 - r}}D\left( r \right)} \cr & = \sum\limits_{r = 2}^5 {\frac{{5!}}{{r!\left( {5 - r} \right)!}}r!\left\{ {1 - \frac{1}{{1!}} + \frac{1}{{2!}} - \frac{1}{{3!}} + ..... + \frac{{{{\left( { - 1} \right)}^r}}}{{r!}}} \right\}} \cr & = 10 + 20 + \left( {60 - 20 + 5} \right) + \left( {60 - 20 + 5 - 1} \right) \cr & = 10 + 20 + 45 + 44 = 119 \cr} $$

Each place can be filled in 3 ways.
So, $$n$$ places can be filled in $$3 \times 3 \times .....\,{\text{to }}n{\text{ times }} = {3^n}.$$
$${3^6} = 729,{3^7} = 2187.$$    So, for $${3^n} \geqslant 900$$   the smallest integral $$n$$ is 7.

The number of students giving exactly one wrong answer $$ = {2^{n - 1}} - {2^{n - 2}}.$$
The number of students giving exactly two wrong answers $$ = {2^{n - 2}} - {2^{n - 3}},\,{\text{e}}{\text{.t}}{\text{.c}}{\text{.}}$$
∴ the total number of wrong answers
$$\eqalign{ & {\text{ = 1}} \cdot \left( {{2^{n - 1}} - {2^{n - 2}}} \right) + 2 \cdot \left( {{2^{n - 2}} - {2^{n - 3}}} \right) + ..... + \left( {n - 1} \right)\left( {2 - {2^0}} \right) + n \cdot 1 = 2047\,\left( {{\text{from question}}} \right) \cr & \therefore \,\,{2^{n - 1}} + {2^{n - 2}} + ..... + 2 - \left( {n - 1} \right) + n = 2047 \cr & {\text{or, }}\frac{{2\left( {{2^{n - 1}} - 1} \right)}}{{2 - 1}} = 4046\,\,\,{\text{or, }}{2^n} = 4048 = {2^{11}}. \cr} $$

The number of selections of $$p$$ things from $$p$$ identical things and $$r - p$$  things from $$q$$ identical things $$ = 1 \times 1.$$

Similarly in all other cases.
∴ the total number of ways
$$\eqalign{ & = p - \left( {r - q} \right) + 1\,\,{\text{or }}q - \left( {r - p} \right) + 1 \cr & = p + q - r + 1. \cr} $$

We know, $$\sum\limits_{r = 1}^n {{r^2} \cdot {\,^n}{C_r}} = n\left( {n - 1} \right){2^{n - 2}} + n \cdot {2^{n - 1}}\,\,\,\,.....\left( 1 \right)$$
$${\text{and }}\sum\limits_{r = 1}^n {{{\left( { - 1} \right)}^{r - 1}} \cdot {r^2} \cdot {\,^n}{C_r}} = 0\,\,\,\,\,.....\left( 2 \right)$$
Adding (1) & (2) we get,
$$\eqalign{ & 2\left[ {{1^2} \cdot {C_1} + {3^2} \cdot {C_3} + {5^2} \cdot {C_5} + .....} \right] = n\left( {n - 1} \right){2^{n - 2}} + n \cdot {2^{n - 1}} \cr & \Rightarrow \left[ {{1^2}{C_1} + {3^2}{C_3} + {5^2}{C_5} + .....} \right] = n\left( {n - 1} \right){2^{n - 3}} + n \cdot {2^{n - 2}}. \cr} $$

No. of triangles $$ = {\,^{2n}}{C_3} - {\,^n}{C_3} - {\,^n}{C_3}$$
$$\eqalign{ & = \frac{{2n\left( {2n - 1} \right)\left( {2n - 2} \right)}}{6} - \frac{{2n\left( {n - 1} \right)\left( {n - 2} \right)}}{6} \cr & = \frac{1}{3}n\left( {n - 1} \right)\left( {3n} \right) = {n^2}\left( {n - 1} \right). \cr} $$

The number of selections of 4 persons including $$A,B = {\,^6}{C_2}.$$
Considering these four as a group, number of arrangements with the other four $$= 5!.$$

But in each group the number of arrangements $$ = 2!\, \times 2!$$
∴ the required number of ways $$ = {\,^6}{C_2} \times 5!\, \times 2!\, \times 2!.$$

Selection of 6 guests $$ = {\,^{10}}{C_6}$$
Permutation of 6 on round table $$= 5!$$
Permutation of 4 on round table $$= 3!$$
Then, total number of arrangements $$ = {\,^{10}}{C_5} \cdot 5! \cdot 3!$$
$$ = \frac{{\left( {10} \right)!}}{{6! \cdot 4!}} \cdot 5! \cdot 3! = \frac{{\left( {10} \right)!}}{{24}}.$$

$$\eqalign{ & 63 = {\,^{2n + 1}}{C_1} + {\,^{2n + 1}}{C_2} + ..... + {\,^{2n + 1}}{C_n} \cr & \therefore \,\,64 = {\,^{2n + 1}}{C_0} + {\,^{2n + 1}}{C_1} + ..... + {\,^{2n + 1}}{C_n} \cr & 64 = \frac{1}{2}\left( {^{2n + 1}{C_0} + {\,^{2n + 1}}{C_1} + ..... + {\,^{2n + 1}}{C_{2n + 1}}} \right) \cr & 64 = \frac{1}{2} \cdot {2^{2n + 1}}. \cr} $$