Permutation and Combination MCQ Questions & Answers in Algebra | Maths

ALLMS
No. of words starting with
$$\eqalign{ & {\text{A}}\,\,:\,\,\underline {\text{A}} \,\,\underline {} \,\,\underline {} \,\,\underline {} \,\,\underline {} \,\,\frac{{4!}}{{2!}} = 12 \cr & {\text{L}}\,\,:\,\,\underline {\text{L}} \,\,\underline {} \,\,\underline {} \,\,\underline {} \,\,\underline {} \,\,4! = 24 \cr & {\text{M}}\,\,:\,\,\underline {\text{M}} \,\,\underline {} \,\,\underline {} \,\,\underline {} \,\,\underline {} \,\,\frac{{4!}}{{2!}} = 12 \cr & {\text{S}}\,\,:\,\,\underline {\text{S}} \,\,\underline {\text{A}} \,\,\underline {} \,\,\underline {} \,\,\underline {} \,\,\frac{{3!}}{{2!}} = 3 \cr & \,\,\,\,\,\,:\,\,\underline {\text{S}} \,\,\underline {\text{L}} \,\,\underline {} \,\,\underline {} \,\,\underline {} \,\,3! = 6 \cr & {\text{SMALL}}\,\,\,®\,\,\,{58^{th}}\,\,{\text{word}} \cr} $$

$$ \times | \times | \times |$$   Crosses can be filled in $$^3{P_3} - {\,^2}{P_2}$$   ways ( $$\because 0$$  cannot go in the first place from the left).
The remaining places can be filled in 3! ways.
∴ the required number of numbers $$ = \left( {^3{P_3} - {\,^2}{P_2}} \right) \times 3!.$$

Since, $$m$$ = number of ways the committee is formed with at least 6 males
$$ = {\,^8}{C_6}.\,{\,^5}{C_5} + {\,^8}{C_7}.\,{\,^5}{C_4} + {\,^8}{C_8}.\,{\,^5}{C_3} = 78$$
and $$n$$ = number of ways the committee is formed with at least 3 females
$$ = {\,^5}{C_3}.\,{\,^8}{C_8} + {\,^5}{C_4}.\,{\,^8}{C_7} + {\,^5}{C_5}.\,{\,^8}{C_6} = 78$$
Hence, $$m = n = 78$$

We know,
$$\eqalign{ & {T_n} = {\,^n}{C_3},{T_{n + 1}} = {\,^{n + 1}}{C_3} \cr & {\text{ATQ, }}{T_{n + 1}} - {T_n} = {\,^{n + 1}}{C_3} - {\,^n}{C_3} = 10 \cr & \Rightarrow \,{\,^n}{C_2} = 10 \cr & \Rightarrow \,\,n = 5. \cr} $$

The number of selections of 1 pair of vowels and 1 pair of consonants $$ = {\,^5}{C_1} \times {\,^{21}}{C_1}.$$
The number of selections of 2 different vowels and 2 different consonants $$ = {\,^5}{C_2} \times {\,^{21}}{C_2}.$$
∴ the required number of words
$$ = {\,^5}{C_1} \times {\,^{21}}{C_1} \times \frac{{4!}}{{2!\,2!}} + {\,^5}{C_2} \times {\,^{21}}{C_2} \times 4!.$$

Six students $${S_1},{S_2},.....,{S_6}$$    can be arranged round a circular table in $$5 !$$ ways. Among these 6 students there are six vactant places, shown by dots $$\left( \bullet \right)$$ in which six teachers can sit in $$6 !$$ ways.
Hence, number of arrangement $$= 5 ! \times 6 !$$

Value $$ = \left( {^{40}{C_9} + {\,^{40}}{C_{10}}} \right) + {\,^{41}}{C_{11}} + ..... + {\,^{50}}{C_{20}}$$
$$\eqalign{ & = \left( {^{41}{C_{10}} + {\,^{41}}{C_{11}}} \right) + {\,^{42}}{C_{12}} + ..... + {\,^{50}}{C_{20}} \cr & = ..... = {\,^{50}}{C_{19}} + {\,^{50}}{C_{20}} = {\,^{51}}{C_{20}}. \cr} $$

The number of numbers with 4 in the middle $${ = ^4}{P_4}{ - ^3}\,{P_3}$$
( $$\because $$  the other four places are to be filled by 0, 1, 2 and 3, and a number cannot begin with 0).
Similarly, the number 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
$$ = \,\left( {^4{P_4} - {\,^3}{P_3}} \right) + \left( {^5{P_4} - {\,^4}{P_3}} \right) + \left( {^6{P_4} - {\,^5}{P_3}} \right) + ..... + \left( {^9{P_4} - {\,^8}{P_3}} \right).$$

Any selection of six digits from 9 digits (excluding 0) will give a number of the required variety.
∴ the required number of numbers $$ = {\,^9}{C_6}.$$

We have to form 7 digit numbers, using the digits 1, 2 and 3 only, such that the sum of the digits in a number = 10.
This can be done by taking 2, 2 , 2, 1, 1, 1, 1, or by taking 2,3,1,1,1,1,1.
∴ Number of ways $$ = \frac{{7!}}{{3!4!}} + \frac{{7!}}{{5!}} = 77.$$