Permutation and Combination MCQ Questions & Answers in Algebra | Maths

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

The number of matches in the first round $$ = {\,^6}{C_2} + {\,^6}{C_2}.$$
The number of matches in the next round $$ = {\,^6}{C_2}.$$
The number of matches in the semifinal round $$ = {\,^4}{C_2}.$$
∴ the required number of matches $$ = {\,^6}{C_2} + {\,^6}{C_2}{ + ^6}{C_2}{ + ^4}{C_2} + 2.$$
Note : For “best of three” at least two matches are played.

The smallest number of people
= total number of possible forecasts
= total number of possible results
$$ = 3 \times 3 \times 3 \times 3 \times 3.$$

Total number of words $$= 7!$$
Out of which $$6!$$ start with $$F$$ and $$6!$$ and with $$E,$$ while $$5!$$ start with $$F$$ and end with $$E.$$

Since, the number of ways to select 2 girls is $$^5{C_2}.$$
Now, 3 boys can be selected in 3 ways.
(1) Selection of $$A$$ and selection of any 2 other boys (except $$B)$$ in $$^5{C_2}$$  ways
(2) Selection of $$B$$ and selection of any 2 two other boys (except $$A)$$ in $$^5{C_2}$$  ways
(3) Selection of 3 boys (except $$A$$ and $$B)$$ in $$^5{C_3}$$  ways
Hence, required number of different teams
$$ = \,{\,^5}{C_2}\left( {^5{C_2}{ + ^5}{C_2}{ + ^5}{C_3}} \right) = 300$$

There are 4 odd digits $$\left( {1,1,3,3} \right)$$   and 4 odd place (first, third, fifth and seventh ). At these places the odd digits can be arranged in $$\frac{{4!}}{{2!2!}} = 6{\text{ ways}}{\text{.}}$$
Then at the remaining 3 places, the remaining three digits $$\left( {2,2,4} \right)$$  can be arranged in $$\frac{{3!}}{{2!}} = 3{\text{ ways}}{\text{.}}$$
$$\therefore $$ The required of number of numbers $$= 6 \times 3 = 18.$$

We have, $$^{10}{C_{n - 1}} > 2.{\,^{10}}{C_n}$$
$$\eqalign{ & \Rightarrow \frac{{10!}}{{\left( {n - 1} \right)!\left( {11 - n} \right)!}} > 2 \cdot \frac{{10!}}{{n!\left( {10 - n} \right)!}} \cr & \Rightarrow \frac{{n!}}{{\left( {n - 1} \right)!}} > 2\frac{{\left( {11 - n} \right)!}}{{\left( {10 - n} \right)!}} \cr & {\text{i}}{\text{.e}}{\text{., }}n > 22 - 2n \cr & \Rightarrow 3n > 22 \cr & \Rightarrow n > \frac{{22}}{3} = 7\frac{1}{3} \cr} $$
$$\therefore $$ Least $$+ ve$$  integral value of $$n = 8.$$

$$\eqalign{ & r \cdot {\,^r}{P_r} = \left( {r + 1 - 1} \right) \times r! = \left( {r + 1} \right)! - r! \cr & \therefore \,\,\sum\limits_{r = 1}^{10} {r \cdot {\,^r}{P_r}} = \sum\limits_{r = 1}^{10} {\left\{ {\left( {r + 1} \right)! - r!} \right\}} \cr & \sum\limits_{r = 1}^{10} {r \cdot {\,^r}{P_r}} = \left( {2! - 1!} \right) + \left( {3! - 2!} \right) + ..... + \left( {11! - 10!} \right) = 11! - 1. \cr} $$

$$\eqalign{ & ^{10}{C_1}{ + ^{10}}{C_2}{ + ^{10}}{C_3}{ + ^{10}}{C_4} \cr & = 10 + 45 + 120 + 210 = 385 \cr} $$

Total number of ways $$ = 6 \times 6 \times .....\,{\text{to }}n{\text{ times }} = {6^n}.$$
Total number of ways to show only even numbers $$ = 3 \times 3 \times .....\,{\text{to }}n{\text{ times }} = {3^n}.$$
∴ the required number of ways $$ = {6^n} - {3^n}.$$