Permutation and Combination MCQ Questions & Answers in Algebra | Maths

$$' - '$$ signs will be put between two $$' + '$$ signs or at the two ends.
There are $$7$$ places for four $$' - '$$ signs. So, the required number of ways
$$ = {\,^7}{C_4} = \frac{{7!}}{{4!\,3!}}$$
(there being no arrangement as the $$' + '$$ signs are identical as well as $$' - '$$ signs are identical).

The candidate is unsuccessful if he fails in 9 or 8 or 7 or 6 or 5 papers.
∴ the number of ways to be unsuccessful
$$\eqalign{ & = {\,^9}{C_9} + {\,^9}{C_8} + {\,^9}{C_7} + {\,^9}{C_6} + {\,^9}{C_5} \cr & = {\,^9}{C_0} + {\,^9}{C_1} + {\,^9}{C_2} + {\,^9}{C_3} + {\,^9}{C_4} \cr & = \frac{1}{2}\left( {^9{C_0} + {\,^9}{C_1} + ..... + {\,^9}{C_9}} \right) = \frac{1}{2} \cdot {2^9} = {2^8}. \cr} $$

The required number of ways $${ = ^{21}}{C_{10}} + {\,^{22}}{C_{10}} + {\,^{23}}{C_{10}} + ..... + {\,^{30}}{C_{10}}$$
$$\eqalign{ & = \left( {^{21}{C_{10}} + {\,^{21}}{C_{11}} + {\,^{22}}{C_{12}} + {\,^{23}}{C_{13}} + ..... + {\,^{30}}{C_{20}}} \right) - {\,^{21}}{C_{10}}\left( {\because \,{\,^n}{C_r} = {\,^n}{C_{n - r}}} \right) \cr & = {\,^{22}}{C_{11}} + {\,^{22}}{C_{12}} + {\,^{23}}{C_{13}} + ..... + {\,^{30}}{C_{20}} - {\,^{21}}{C_{10}}\left( {\because \,{\,^n}{C_{r - 1}} + {\,^n}{C_r} = {\,^{n + 1}}{C_r}} \right) \cr & ......................................................................... \cr & = {\,^{31}}{C_{20}} - {\,^{21}}{C_{10}}. \cr} $$

Alphabetical order is
A, C, H, I, N, S
No. of words starting with A - 5!
No. of words starting with C - 5!
No. of words starting with H - 5!
No. of words starting with I - 5!
No. of words starting with N - 5!
SACHIN - 1
∴ sachin appears at serial no 601

As $$0 < x < 1 ,$$   we have $$p < q$$
The number of rational numbers $$= 5 + 4 + 3 + 2 + 1 = 15.$$
When $$p, q$$  have a common factor, we get some rational numbers which are not different from those already counted. There are 4 such numbers :
$$\frac{2}{4},\frac{2}{6},\frac{3}{6},\frac{4}{6}$$
Therefore, required number of rational numbers $$= 15 - 4 = 11.$$

The given situation in statement - 1 is equivalent to find the non-negative integral solutions of the equation
$${x_1} + {x_2} + {x_3} + {x_4} + {x_5} = 6$$
which is co-eff. of $${x^6}$$  in the expansion of
$$\eqalign{ & {\left( {1 + x + {x^2} + {x^3} + ......\infty } \right)^5} = {\text{co-eff}}{\text{. of }}{x^6}{\text{ in }}{\left( {1 - x} \right)^{ - 5}} \cr & = {\text{co-eff}}{\text{. of }}{x^6}{\text{ in}}\,\,1 + 5x + \frac{{5.6}}{{2!}}{x^2}...... \cr & = \frac{{5.6.7.8.9.10}}{{6!}} = \frac{{10!}}{{6!4!}} = {\,^{10}}{C_6} \cr} $$
∴ Statement - 1 is wrong.
Number of ways of arranging $$6A’s$$  and $$4B’s$$  in a row
$$ = \frac{{10!}}{{6!4!}} = {\,^{10}}{C_6}$$    which is same as the number of ways the child can buy six ice-creams.
∴ Statement - 2 is true.

Total number of triplets without restriction $$ = n \times n \times n.$$
The number of triplets with all different co-ordinates $$ = {\,^n}{P_3}.$$
∴ the required number of triplets $$ = {n^3} - n\left( {n - 1} \right)\left( {n - 2} \right).$$

From the question,
$$\eqalign{ & ^n{P_{n - 1}} = k \cdot \left\{ {^1{C_1} \times {\,^{n - 2}}{C_{n - 3}} \times \frac{{\left( {n - 1} \right)!}}{{2!}} + {\,^{n - 1}}{P_{n - 1}}} \right\} \cr & \Rightarrow \,\,n!\, = k\left\{ {\frac{{\left( {n - 2} \right) \times \left( {n - 1} \right)!}}{2} + \left( {n - 1} \right)!} \right\} \cr & {\text{or, }}n = k\left\{ {\frac{{n - 2}}{2} + 1} \right\}. \cr} $$

If $$n$$ is odd, $${3^n} = 4{\lambda _1} - 1,{5^n} = 4{\lambda _2} + 1$$
$$ \Rightarrow {2^n} + {3^n} + {5^n}$$   is not divisible by 4, as $${2^n} + {3^n} + {5^n}$$   will be in the form of $$4\lambda + 2.$$
Thus, total number of ways of selecting $$'n'$$ is equal to 49.

$$\eqalign{ & {\text{Set, }}S = \left\{ {1,2,3,.....,12} \right\} \cr & A \cup B \cup C = S,A \cap B = B \cap C = A \cap C = \phi \cr} $$
$$\therefore $$ The number of ways to partition
$$\eqalign{ & = {\,^{12}}{C_4} \times {\,^8}{C_4} \times {\,^4}{C_4} \cr & = \frac{{12!}}{{4!8!}} \times \frac{{8!}}{{4!4!}} \times \frac{{4!}}{{4!0!}} = \frac{{12!}}{{{{\left( {4!} \right)}^3}}} \cr} $$