Permutation and Combination MCQ Questions & Answers in Algebra | Maths

The number of triangles with vertices on different lines
$$ = {\,^p}{C_1} \times {\,^p}{C_1} \times {\,^p}{C_1} = {p^3}.$$
The number of triangles with 2 vertices on one line and the third vertex on any one of the other two lines
$$ = {\,^3}{C_1}\left\{ {^p{C_2} \times {\,^{2p}}{C_1}} \right\} = 6p \cdot \frac{{p\left( {p - 1} \right)}}{2}$$
∴ the required number of triangles $$ = {p^3} + 3{p^2}\left( {p - 1} \right).$$
Note : The word “maximum” ensures that no selection of points from each of the three lines are collinear.

\[p:\begin{array}{*{20}{c}} {TTH}&{TH}&H&T&0&{{\rm{place}}}\\ 5&4&3&2&1&{{\rm{ways}}} \end{array}\]
Total no. of ways $$= 5! = 120$$
Since all numbers are $$> 20,000$$
$$\therefore $$ all numbers 2, 3, 5, 7, 9 can come at first place.
\[q:\begin{array}{*{20}{c}} {TTH}&{TH}&H&T&0&{{\rm{place}}}\\ 3&4&3&2&1&{{\rm{ways}}} \end{array}\]
Total no. of ways $$= 3 × 4! = 72$$
($$\because $$ 2 and 9 can not be put at first place)
So, $$p : q = 120 : 72 = 5 : 3$$

∴ Required number of ways $$ =\,\, ^6{C_4} \times {\,^3}{C_1} \times 4!$$
$$ = 15 \times 3 \times 24 = 1080$$

\[\begin{gathered} \begin{array}{*{20}{c}} {Possibilities}&{Selections}&{Arrangements} \\ {{\text{One triplet, two different}}}&{^3{C_1} \times {\,^2}{C_2}}&{^3{C_1} \times {\,^2}{C_2} \times \frac{{5!}}{{3!}} = 60} \\ {{\text{Two pairs, one different}}}&{^3{C_2} \times {\,^1}{C_1}}&{\underline {^3{C_2} \times {\,^1}{C_1} \times \frac{{5!}}{{2!\,2!}} = 90} } \end{array} \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\therefore \,\,{\text{the required number of ways}} = 150. \hfill \\ \end{gathered} \]

If 0 is placed in the units place of the upper number then the units place of the lower number can be filled in 10 ways (filling by any one of 0, 1, 2, . . . . . , 9).
If 1 is placed in the units place of the upper number then the units place of the lower number can be filled in 9 ways (filling by any one of 0, 1, 2, . . . . . , 8), e.t.c.
∴ the units column can be filled in $$10 + 9 + 8 + . . . . . + 1,$$    i.e., 55 ways.
Similarly for the second and the third columns. The number of ways for the fourth column $$= 8 + 7 + . . . . . + 1 = 36$$
∴ the required number of ways $$ = 55 \times 55 \times 55 \times 36.$$

30 marks to be alloted to 8 questions. Each question has to be given $$ \geqslant 2$$  marks
Let questions be $$a, b, c, d, e, f, g, h$$
and $$a + b + c + d + e + f + g + h = 30$$
Let, $$a = {a_1} + 2{\text{ so, }}{a_1} \geqslant 0$$
$$\eqalign{ & b = {a_2} + 2{\text{ so, }}{a_2} \geqslant 0,.....,{a_8} \geqslant 0 \cr & {\text{So, }}\left. {{a_1} + {a_2} + ..... + {a_8} + 2 + 2 + ..... + 2} \right\} = 30 \cr & \Rightarrow {a_1} + {a_2} + ..... + {a_8} = 30 - 16 = 14. \cr} $$
So, this is a problem of distributing 14 articles in 8 groups.
Number of ways $$ = {\,^{14 + 8 - 1}}{C_{8 - 1}} = {\,^{21}}{C_7}$$

The number of ways to fill the three even places by 4 consonants $$ = {\,^4}{P_3}.$$
After filling the even places, remaining places can be filled in $${\,^4}{P_4}$$  ways.
So, the required number of words $$ = {\,^4}{P_3} \times {\,^4}{P_4}.$$

Required sum
$$= (2 + 4 + 6 + . . . + 100) + (5 + 10 + 15 + . . . + 100) - (10 + 20 + . . . + 100)$$
$$= 2550 + 1050 $$-$$ 530 = 3050.$$

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

The letter of word $$COCHIN$$   in alphabetic order are $$C, C, H, I, N, O.$$
Fixing first letter $$C$$ and keeping $$C$$ at second place, rest 4 can be arranged in 4! ways.
Similarly the words starting with $$CH, CI, CN$$   are 4! in each case.
Then fixing first two letters as $$CO$$  next four places when filled in alphabetic order give the word $$COCHIN.$$
∴ Numbers of words coming before $$COCHIN$$   are
\[4 \times 4! = 4 \times 24 = 96\]