Permutation and Combination MCQ Questions & Answers in Algebra | Maths

We have, $$S = \sum\limits_{k = 0}^{n - 1} {^{k + 2}{P_2}} = 2!\sum\limits_{k = 0}^{n - 1} {^{k + 2}{C_2}} $$
$$\eqalign{ & = \left( 2 \right)\left[ {^2{C_2} + {\,^3}{C_2} + {\,^4}{C_2} + ..... + {\,^{n + 1}}{C_2}} \right] \cr & {\text{But,}}{{\text{ }}^2}{C_2} + {\,^3}{C_2} + {\,^4}{C_2} + ..... + {\,^{n + 1}}{C_2} \cr & = \left( {^3{C_3} + {\,^3}{C_2}} \right) + {\,^4}{C_2} + ..... + {\,^{n + 1}}{C_2} \cr & = \left( {^4{C_3} + {\,^4}{C_2}} \right) + {\,^5}{C_2} + ..... + {\,^{n + 1}}{C_2} \cr & = {\,^5}{C_3} + {\,^5}{C_2} + ..... + {\,^{n + 1}}{C_2} \cr & = {\,^{n + 2}}{C_3} = \frac{1}{6}n\left( {n + 1} \right)\left( {n + 2} \right) \cr & \Rightarrow 3S = n\left( {n + 1} \right)\left( {n + 2} \right) \cr} $$
$$ \Rightarrow n$$  divides $$3S,\left( {n + 1} \right)$$   divides $$3S$$  and $$\left( {n + 2} \right)$$  divides $$3S.$$

$$\because $$ $$r, s, t$$  are prime numbers,
∴ Section of $$(p, q)$$  can be done as follows
∴ $$r$$ can be selected 1 + 1 + 3 = 5 ways
\[\begin{array}{l} \begin{array}{*{20}{c}} p\\ {{r^0}}\\ {{r^1}} \end{array}\,\,\,\,\,\,\begin{array}{*{20}{c}} q\\ {{r^2}}\\ {{r^2}} \end{array}\\ \begin{array}{*{20}{c}} {{r^2}}&{{r^0},{r^1},{r^2}} \end{array} \end{array}\]
Similarly $$s$$ and $$t$$ can be selected in 9 and 5 ways respectively .
∴ Total ways $$ = 5 \times 9 \times 5 = 225$$

First we have to select 2 men for bow side and 3 for stroke side. The number of selections of the crew for two sides $$ = {\,^5}{C_2} \times {\,^3}{C_3}$$
For each selection there are 4 persons on both sides, who can be arranged in $$4! \times 4!$$  ways. Required number of arrangement
$$ = {\,^5}{C_2} \times {\,^3}{C_3} \times 4!\, \times 4! = 5760$$

Since as per the give condition $$x > - 1,$$   so $$x$$ is non negative integer, $$y > - 2$$   so $$y = - 1 + b$$   and similarly $$z > 3$$  so $$z = - 2 + c$$
or, $$\left( x \right) + \left( { - 1 + b} \right) + \left( { - 2 + c} \right) = 23$$
or, $$x + b + c = 23$$
and we need to find the number of non negative integral solution of the equation $$x + b + c = 23$$    which is,
$$^{23 + 3 - 1}{C_{3 - 1}} = {\,^{25}}{C_2} = {\,^{25}}{C_{23}}$$

Sum of $$7$$ digits $$= a$$  multiple of $$9$$ Since sum of number $$1, 2, 3, . . . . . , 8, 9$$    is $$45\,$$ (Since a number is divisible by $$9$$ if sum of its digits is divisible by $$9.$$ ) So, two left number should also have sum as $$9.$$ The pairs to be left are $$\left( {1,8} \right),\left( {2,7} \right),\left( {3,6} \right),\left( {4,5} \right).$$      With each pair left, number of $$7$$ digit numbers $$= 7!.$$  So, with all $$4$$ pairs total seven digits number $$= 4 × 7!$$

As for given question two cases are possible.
(i) Selecting 4 out of first five question and 6 out of remaining 8 question $$ = {\,^5}{C_4} \times {\,^8}{C_6} = 140$$     choices.
(ii) Selecting 5 out of first five question and 5 out of remaining 8 questions $$ = {\,^5}{C_5} \times {\,^8}{C_5} = 56$$     choices.
∴ total number of choices = 140 + 56 = 196.

The number of committees of $$4$$ gentlemen $$ = {\,^4}{C_4} = 1.$$
The number of committees of $$3$$ gentlemen, $$1$$ wife $$ = {\,^4}{C_3} \times {\,^1}{C_1}$$
($$\because $$ after selecting $$3$$ gentlemen only $$1$$ wife is left who can be included).
The number of committees of $$2$$ gentlemen, $$2$$ wives $$ = {\,^4}{C_2} \times {\,^2}{C_2}.$$
The number of committees of $$1$$ gentleman, $$3$$ wives $$ = {\,^4}{C_1} \times {\,^3}{C_3}.$$
The number of committees of $$4$$ wives $$= 1.$$
∴ the required number of committees $$= 1 + 4 + 6 + 4 + 1 = 16.$$

As 5 are always to be excluded and 6 always to be included, 5 players to be chosen from 14. Hence, required number of ways are $$^{14}{C_5} = 2002.$$

Ten pearls of one colour can be arranged in $$\frac{1}{2} \cdot \left( {10 - 1} \right)!{\text{ ways}}{\text{.}}$$    The number of arrangements of 10 pearls of the other colour in 10 places between the pearls of the first colour $$= 10!$$
$$\therefore $$ Required number of ways $$ = \frac{1}{2} \times 9!\, \times 10! = 5{\left( {9!} \right)^2}$$

(i) Number of 3 digit nos. $$= 7 \times 6 \times 5 = 210$$
(ii) Number of 2 digit nos. $$ = 7 \times 6 = 42$$
(iii) Number of 1 digit nos. $$= 7$$
Total number of nos. $$= 210 + 42 + 7 = 259$$