Permutation and Combination MCQ Questions & Answers in Algebra | Maths

Any selection of four digits from the ten digits 0, 1, 2, 3, . . . . . , 9 gives one number. So, the required number of numbers $$ = {\,^{10}}{C_4}.$$

Total number of permutations $$ = \frac{{9!}}{{2!}}$$
Number of those containing $$'HIN' = 7!$$
Number of those containing $$'DUS' = \frac{{7!}}{{2!}}$$
Number of those containing $$'TAN' = 7!$$
Number of those containing $$'HIN'$$  and $$'DUS' = 5!$$
Number of those containing $$'HIN'$$  and $$'TAN' = 5!$$
Number of those containing $$'TAN'$$  and $$'DUS' = 5!$$
Number of those containing $$'HIN', 'DUS'$$   and $$'TAN' = 3!$$
Required number
$$ = \frac{{9!}}{{2!}}\left( {7!\, + 7!\, + \frac{{7!}}{2}} \right) + 3 \times 5!\, - 3! = 169194.$$

$$a + a + d + a + 2d = 21$$     or, $$a + d = 7$$
∴ $$a + c = 14$$   and $$b = 7.$$
The number of positive integral solutions of $$(a + c = 14)$$   is 13.

Leaving the ground floor and second floor, their are 10 floors in which three groups of people can left the lift cabin in $$^{10}{P_3}$$  ways, i.e. 720 ways.

\[\begin{gathered} \underline {\begin{array}{*{20}{c}} {Possibilities}&{Selection} \\ \begin{gathered} {\text{At least one rupee coin, any number of 50 - paisa coins,}} \hfill \\ {\text{ any number of 25 - paisa coins}} \hfill \\ \end{gathered} &{2 \times 4 \times 5} \\ {{\text{At least two 50 - paisa coins, any number of 25 - paisa coins}}}&{2 \times 5} \\ {{\text{One 50 - paisa coin, at least two 25 - paisa coins}}}&{1 \times 3} \\ {{\text{Four 25 - paisa coins}}}&1 \end{array}} \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\therefore \,\,{\text{the required number of ways}} = {\text{54}} \hfill \\ \end{gathered} \]

We have,
$$\eqalign{ & S = \sum\limits_{k = 1}^n {k\left( {k!} \right)} = \sum\limits_{k = 1}^n {\left\{ {\left( {k + 1} \right) - 1} \right\}\left( {k!} \right)} \cr & = \sum\limits_{k = 1}^n {\left\{ {\left( {k + 1} \right)! - k!} \right\} = \left( {n + 1} \right)! - 1} \cr & \Rightarrow S + 1 = \left( {n + 1} \right)! \cr & {\text{Thus, }}\frac{{S + 1}}{{n!}} \in {\text{integer}}{\text{.}} \cr} $$

We know that the number of ways of distributing $$n$$ identical items among $$r$$ persons, when each one of them receives at least one item is $$^{n - 1}{C_{r - 1}}$$
∴ The required number of ways
$${ = ^{8 - 1}}{C_{3 - 1}} = {\,^7}{C_2} = \frac{{7!}}{{2!5!}} = \frac{{7 \times 6}}{{2 \times 1}} = 21$$

$$\frac{{1,2,3,4}}{{ \times \times \times \times }}\,\,5\,\,\frac{{6,7,8,9}}{{ \times \times \times \times }}$$     The required number of numbers $$ = {\,^4}{P_4} \times {\,^4}{P_4}.$$

$$X - X - X - X - X.$$    The four digits 3, 3, 5, 5 can be arranged at $$(-)$$ places in $$\frac{{4!}}{{2!2!}} = 6\,\,{\text{ways}}{\text{.}}$$
The five digits 2, 2, 8, 8, 8 can be arranged at $$(X)$$ places in $$\frac{{5!}}{{2!3!}} = 10\,\,{\text{ways}}{\text{.}}$$
Total no. of arrangements $${\text{ = }}\,{\text{6}} \times {\text{10}}\,{\text{ = }}\,{\text{60}}\,\,{\text{ways}}{\text{.}}$$

We have in all 12 points. Since, 3 points are used to form a triangle, therefore the total number of triangles including the triangles formed by collinear points on $$AB, BC$$   and $$CA$$  is $$^{12}{C_3} = 220.$$   But this includes the following :
The number of triangles formed by 3 points on $$AB = {\,^3}{C_3} = 1.$$
The number of triangles formed by 4 points on $$BC = {\,^4}{C_3} = 4.$$
The number of triangles formed by 5 points on $$CA = {\,^5}{C_3} = 10.$$
Hence, required number of triangles $$ = 220 - \left( {10 + 4 + 1} \right) = 205.$$