Permutation and Combination MCQ Questions & Answers in Algebra | Maths

Out of 10 points let n points are collinear.
Then the number of triangles is
$$\eqalign{ & {}^{10}{C_3} - {}^n{C_3} = 110 \cr & \Rightarrow \frac{{10 \times 9 \times 8}}{6} - \frac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{6} = 110 \cr & \Rightarrow n\left( {n - 1} \right)\left( {n - 2} \right) = 60 \cr & \Rightarrow n = 5 \cr} $$

$${2^p} \cdot {6^q} \cdot {15^r} = {2^{p + q}} \cdot {3^{q + r}} \cdot {5^r}$$
∴ the number of proper divisors
= {total number of selections from $$\left( {p + q} \right)$$  twos, $$\left( {q + r} \right)$$  threes and $$r$$ fives} $$- 2$$
$$ = \left( {p + q + 1} \right)\left( {q + r + 1} \right)\left( {r + 1} \right) - 2.$$

(The number of ways of dividing in 3 equal groups) $$ \times \left( {3!} \right)$$
= the number of ways to distribute equally among 3 persons.

Since the largest digit is in the middle, the middle digit is greater than or equal to 4 the number of numbers with 4 in the middle $$ = {\,^4}{P_4} - {\,^3}{P_3}.$$
(∵ the other four places are to be filled by 0, 1, 2 and 3 and a number cannot begin with 0). Similarly, the numbers of numbers with 5 in the middle $$ = {\,^5}{P_4} - {\,^4}{P_3},{\text{ e}}{\text{.t}}{\text{.c}}{\text{.)}}$$
∴ The required number of numbers
$$\eqalign{ & = \left( {^4{P_4} - {\,^3}{P_3}} \right) + \left( {^5{P_4} - {\,^4}{P_3}} \right) + \left( {^6{P_4} - {\,^5}{P_3}} \right) + ..... + \left( {^9{P_4} - {\,^8}{P_3}} \right) \cr & = \sum\limits_{n = 4}^9 {^n{P_4} - } \sum\limits_{n = 3}^8 {^n{P_3}} \cr} $$

Let the number be $$n = pqr.$$   As $$p + q + r$$   is even, there are following cases:
$$\left( {\text{i}} \right)p,q,r$$   all are even $$ \to 4 \cdot 5 \cdot 5 = 100{\text{ ways}}$$
$$\left( {\text{ii}} \right)p$$  even and $$q,r$$  are odd $$ \to 4 \cdot 5 \cdot 5 = 100{\text{ ways}}$$
$$\left( {\text{iii}} \right)p$$  odd, $$q$$ and $$r$$ are even $$ \to 5 \cdot 5 \cdot 5 = 125{\text{ ways}}$$
$$\left( {\text{iv}} \right)p$$  odd, $$q$$ even, $$r$$ odd $$ \to 5 \cdot 5 \cdot 5 = 125{\text{ ways}}$$
Total ways $$= 100 + 100 + 125 + 125 = 450.$$

The number of positive integral solutions
= co-efficient of $${x^n}\,{\text{in }}{\left( {x + {x^2} + {x^3} + .....} \right)^3}$$
= co-efficient of $${x^{n - 3}}\,{\text{in }}{\left( {1 + x + {x^2} + .....} \right)^3}$$
= co-efficient of $${x^{n - 3}}\,{\text{in }}{\left( {1 - x} \right)^{ - 3}}$$
= co-efficient of $${x^{n - 3}}\,{\text{in }}\left\{ {^2{C_0} + {\,^3}{C_1}x + {\,^4}{C_2}{x^2} + .....} \right\}$$
$$ = {\,^{n - 1}}{C_{n - 3}} = \frac{{\left( {n - 1} \right)!}}{{\left( {n - 3} \right)!\,\,2!}} = \frac{{\left( {n - 1} \right)\left( {n - 2} \right)}}{2}.$$

A triangle can be formed by using three non-collinear points.
So, the number of triangles formed by 18 non-collinear points $$ = {\,^{18}}{C_3}$$
But according to the question, 5 points are collinear.
Hence, exact number of triangles
$$\eqalign{ & = {\,^{18}}{C_3} - {\,^5}{C_3} = \frac{{18!}}{{3!15!}} - \frac{{5!}}{{3!2!}} \cr & = \frac{{16 \times 17 \times 18}}{{2 \times 3}} - \frac{{4 \times 5}}{2} = 816 - 10 = 806. \cr} $$

Ways: \[\begin{array}{*{20}{c}} \times \\ 9 \end{array}\,\,\begin{array}{*{20}{c}} \times \\ {10} \end{array}\,\,\begin{array}{*{20}{c}} \times \\ {10} \end{array}\,\,\begin{array}{*{20}{c}} \times \\ {10} \end{array}\,\,\begin{array}{*{20}{c}} \times \\ {10} \end{array}\,\,\begin{array}{*{20}{c}} \times \\ 6 \end{array}\]
0 cannot go in the first place. In other four places any digit can go. After filling the first five places, the last place can be filled by 0 or 5, 1 or 6, 2 or 7, 3 or 8, 4 or 9 depending upon whether the sum of five digits filled is of the form $$5m, 5m + 4, 5m + 3, 5m + 2$$      or $$5m + 1$$  respectively.
∴ the required number of numbers $$ = 9 \times {10^4} \times 2.$$

Ways: \[\begin{array}{*{20}{c}} \times \\ 9 \end{array}\,\,\begin{array}{*{20}{c}} \times \\ 9 \end{array}\,\,\begin{array}{*{20}{c}} \times \\ 9 \end{array}\,\,\begin{array}{*{20}{c}} \times \\ 9 \end{array}\,\,\begin{array}{*{20}{c}} \times \\ 9 \end{array}\]
0 cannot be placed in the first place. In the next place any digit except the one used in the first place can be used, e.t.c.

No. of ways getting one correct
$$ = {\,^7}{C_1} \cdot 6!\left( {1 - \frac{1}{{1!}} + \frac{1}{{2!}} - \frac{1}{{3!}} + ..... + \frac{1}{{6!}}} \right) = {\,^7}{C_1} \cdot 265$$
No. of ways getting two correct
$$ = {\,^7}{C_2} \cdot 5!\left( {1 - \frac{1}{{1!}} + \frac{1}{{2!}} - \frac{1}{{3!}} + ..... - \frac{1}{{5!}}} \right) = {\,^7}{C_2} \cdot 44$$
No. of ways getting three correct
$$ = {\,^7}{C_3} \cdot 4!\left( {1 - \frac{1}{{1!}} + \frac{1}{{2!}} - \frac{1}{{3!}} + \frac{1}{{4!}}} \right) = {\,^7}{C_3} \cdot 9$$
Required no. of ways
$$ = {\,^7}{C_3} \cdot 9 + {\,^7}{C_2} \cdot 44 + {\,^7}{C_1} \cdot 265.$$