Permutation and Combination MCQ Questions & Answers in Algebra | Maths

Four digits number can be arranged in $$3 \times 4!$$  ways.
Five digits number can be arranged in 5! ways.
Number of integers $$ = 3 \times 4!\,\, + 5! = 192.$$

Total number of ways $$ = m \times m \times .....\,{\text{to }}n\,{\text{times = }}{m^n}.$$
The number of ways in which one gets all the prizes $$= m.$$
∴ the required number of ways $$ = {m^n} - m.$$

Each element of the set $$A$$ can be given the image in the set $$A$$ in $$k$$ ways.
∴ the required number of functions, i.e., $$n\left( S \right) = k \times k \times .....\left( {k\,{\text{times}}} \right) = {k^k}.$$

A couple and $$6$$ guests can be arranged in $$(7 - 1) !$$  ways. But the two people forming the couple can be arranged among themselves in $$2 !$$ ways.
∴ the required number of ways $$ = 6!\, \times 2!.$$

$$1008 = {2^4} \times {3^2} \times 7$$
∴ the required number of even proper divisors
= total number of selections of at least one 2 and any number of $$3’s$$ or $$7’s$$
$$ = 4 \times \left( {2 + 1} \right) \times \left( {1 + 1} \right) - 1.$$

As for given question two cases are possible.
(i) Selecting 4 out of first five question and 6 out of remaining 8 question $$ = {\,^5}{C_4} \times {\,^8}{C_6} = 140{\text{ choices}}{\text{.}}$$
(ii) Selecting 5 out of first five question and 5 out of remaining 8 questions $$ = {\,^5}{C_5} \times {\,^8}{C_5} = 56{\text{ choices}}{\text{.}}$$
Therefore, total number of choices $$ = 140 + 56 = 196.$$

If we see the blocks in terms of lines then there are $$2m$$ vertical lines and $$2n$$ horizontal lines. To form the required rectangle we must select two horizontal lines, one even numbered (out of 2, 4, . . . . . $$2n$$) and one odd numbered (out of 1, 3, . . . . . $$2n-1$$ ) and similarly two vertical lines. The number of rectangles is
$$^m{C_1}.{\,^m}{C_1}.{\,^n}{C_1}.{\,^n}{C_1} = {m^2}{n^2}$$

As the order of $$A, B, C$$  is not to change they are to be treated identical in arrangement. So, the required number of ways $$ = \frac{{7!}}{{3!}}.$$

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

Total number of words that can be formed using 5 letters out of 10 given different letters
$$ = 10 \times 10 \times 10 \times 10 \times 10$$     (as letters can repeat )
= 1,00,000
Number of words that can be formed using 5 different letters out of 10 different letters
$$ = {\,^{10}}{P_5}$$  (none can repeat)
$$ = \frac{{10!}}{{5!}} = 30,240$$
∴ Number of words in which at least one letter is repeated
= total words - words with none of the letters repeated
$$= 1,00,000 - 30,240 = 69760$$