Permutation and Combination MCQ Questions & Answers in Algebra | Maths

The required number of ways
= total number of ways of selecting any number of books from $$m$$ different books, any number of pens from $$n$$ different pens and any number of pencils from $$p$$ different pencils $$- 1$$
$$\eqalign{ & = \left( {^m{C_0} + {\,^m}{C_1} + ..... + {\,^m}{C_m}} \right)\left( {^n{C_0} + {\,^n}{C_1} + ..... + {\,^n}{C_n}} \right) \times \left( {^p{C_0} + {\,^p}{C_1} + ..... + {\,^p}{C_p}} \right) - 1 \cr & = {2^m} \times {2^n} \times {2^p} - 1. \cr} $$

No. of ways in which 6 men can be arranged at a round table $$= (6 - 1)! = 5!$$
Now women can be arranged in $$^6{P_5} = 6!$$   Ways.
Total Number of ways = $$6!\,\, \times \,\,5!$$

Number of ways of displaying without restriction $$= 3^5$$
Number of ways in which all places are occupied by single variety $$= 3 \times 1$$
Number of ways in which all places are occupied by two different varieties $$ = {\,^3}{C_2}\left[ {{2^5} - 2} \right]$$
[$$\because $$ There are two ways when all places will be occupied by single variety each.]
$$\therefore $$ No. of way of displaying all three varieties
$$ = {3^5} - 3 - {\,^3}{C_2}\left( {{2^5} - 2} \right) = 150$$

$$\eqalign{ & ^n{C_r} + 4 \cdot {\,^n}{C_{r - 1}} + 6 \cdot {\,^n}{C_{r - 2}} + 4 \cdot {\,^n}{C_{r - 3}} + {\,^n}{C_{r - 4}} \cr & = \left( {^n{C_r} + {\,^n}{C_{r - 1}}} \right) + 3\left( {^n{C_{r - 1}} + {\,^n}{C_{r - 2}}} \right) + 3\left( {^n{C_{r - 2}} + {\,^n}{C_{r - 3}}} \right) + \left( {^n{C_{r - 3}} + {\,^n}{C_{r - 4}}} \right) \cr & = {\,^{n + 1}}{C_r} + 3 \cdot {\,^{n + 1}}{C_{r - 1}} + 3 \cdot {\,^{n + 1}}{C_{r - 2}} + {\,^{n + 1}}{C_{r - 3}} \cr & = \left( {^{n + 1}{C_r} + {\,^{n + 1}}{C_{r - 1}}} \right) + 2\left( {^{n + 1}{C_{r - 1}} + {\,^{n + 1}}{C_{r - 2}}} \right) + \left( {^{n + 1}{C_{r - 2}} + {\,^{n + 1}}{C_{r - 3}}} \right) \cr & = {\,^{n + 2}}{C_r} + 2 \cdot \left( {^{n + 2}{C_{r - 1}} + {\,^{n + 1}}{C_{r - 2}}} \right) \cr & = \left( {^{n + 2}{C_r} + {\,^{n + 2}}{C_{r - 1}}} \right) + \left( {^{n + 2}{C_{r - 1}} + {\,^{n + 2}}{C_{r - 2}}} \right) \cr & = {\,^{n + 3}}{C_r} + {\,^{n + 3}}{C_{r - 1}} = {\,^{n + 4}}{C_r} \cr} $$

Between $$m$$ and $$n,$$ there are $$n - m - 1$$   elements. Each subset contains $$m$$ and $$n$$ and for all of other $$n - m - 1$$   elements, there are two possibilities so, no. of subset $$ = {2^{n - m - 1}}.$$

The number of ways of selecting 3 persons from 12 people under the given conditon :
= Number of ways of arranging 3 people among 9 people seated in a row, so that no two of them are consecutive
= Number of ways of choosing 3 places out of the 10 [8 in between and 2 extremes]
$$ = {\,^{10}}{C_3} = \frac{{10 \times 9 \times 8}}{{3 \times 2 \times 1}} = 5 \times 3 \times 8 = 120$$

The number of words like $$C - C = \frac{{7!}}{{2!\,2!}} = $$   the number of words like $$L - L.$$
The number of words beginning or ending with $$C,L = 2 \times \frac{{7!}}{{2!\,}}.$$
The number of words beginning or ending with $$\left( {C\,{\text{or }}L} \right),$$   $$T = 2\left( {2!} \right) \times \frac{{7!}}{{2!\,2!}}.$$
∴ the required number of words
$$\eqalign{ & = \frac{{7!}}{{2!\,2!}} + \frac{{7!}}{{2!\,2!}} + 2 \times \frac{{7!}}{{2!}} + 2\left( {2!} \right) \times \frac{{7!}}{{2!\,2!}} \cr & = \left( {\frac{1}{4} + \frac{1}{4} + 1 + 1} \right)\left( {7!} \right). \cr} $$

Number of required triangles
$$\eqalign{ & = \,{\,^{10}}{C_3} - {\,^6}{C_3} \cr & \frac{{10 \times 9 \times 8}}{6} - \frac{{6 \times 5 \times 4}}{6} \cr & = 120 - 20 \cr & = 100 \cr} $$

The number of points of intersection of 37 straight lines is $$^{37}{C_2}.$$  But 13 of them pass through the point $$A.$$ Therefore instead of getting $$^{13}{C_2}.$$  points we get merely one point. Similary 11 straight lines out of the given 37 straight lines intersect at $$B.$$ Therefore instead of getting $$^{11}{C_2}.$$  points, we get only one point. Hence, the number of intersection points of the lines is
$$^{37}{C_2} - {\,^{13}}{C_2} - {\,^{11}}{C_2} + 2 = 535.$$

From every arrangement of 7 letters we get a pair by putting a sign (,) after the four letters from the left.
∴ the required number of pairs $$ = \frac{{7!}}{{2!\,2!}}.$$