58 questions · self-marked practice — reveal the answer and mark yourself.
= 6 × 10
= 60
Number of outcomes $=2^6$
This number includes one case when the student solves NONE of the questions.
$\therefore$ Required number of ways $=2^6-1=64-1=63$
Number of instances $=2^{12}$
This number includes one case in when all 12 lamps are OFF.
$\therefore$ Required Number of ways $=2^{12}-1=4095$
This number includes one case when the student passes in all subjects. Required number of ways = 128 – 1 = 127
= 11 × 10 × 9 = 990
${ }^{20} C _{16}-{ }^{19} C _{16}$
${ }^{15} C_4+{ }^{15} C_5$
${ }^{80} C_2$
$=40 \times 79=3160$
${ }^{15} C _4$
$\therefore$ Required number of arrangements $=12 ! \times{ }^{13} P _2$
= 12! × 13 × 12
= 13 × 12! × 12
= 12 × 13!
$\therefore$ Required number of arrangements $=\frac{9 !}{4 !}$
No two women should sit together.
There are 8 gaps created by 8 men’s seats.
$\therefore$ Women can be seated in 8 gaps in ${ }^8 P _6$ ways.
$\therefore$ Required number of arrangements $=7 ! \times{ }^8 P _6$
∴ They can be arranged in(15 – 1)! = 14! ways.
But, they should not have the same neighbour in any two arrangements.
Around the table, arrangements (i.e., clockwise and anticlockwise) coincide.
$\therefore$ Required number of arrangements $=\frac{14 !}{2}$
∴ 8 friends can sit around a table in 7!
= 7 × 6 × 5 × 4 × 3 × 2 × 1
= 5040 ways.
Number of such numbers $=\frac{5 !}{2 !}+\frac{5 !}{2 !}=5 !=120$
Case II: Numbers are formed from 2, 2, 3, 3 and any one of 4 or 5
Number of such numbers $=\frac{5 !}{2 ! 2 !}+\frac{5 !}{2 ! 2 !}=60$
Required number of numbers = 120 + 60 = 180
$\therefore$ Number of ways to arrange the letters of word ARRANGE $=\frac{7 !}{2 ! 2 !}=1260$
Consider the words in which 2A are together and 2R are together. Let us consider 2A as one unit and 2R as one unit.
These two units with remaining 3 letters can be arranged in $=\frac{5 !}{2 ! 2 !}=30$ ways.
Number of arrangements in which neither 2A together nor 2R are together = 1260 – 30 = 1230
$\therefore$ Required number of arrangements $=\frac{10 !}{2 ! 2 ! 3 !}$
$\frac{1}{(n-1) !}+\frac{1-n}{(n+1) !}$
$\frac{n+2}{n !}-\frac{3 n+1}{(n+1) !}$
$\frac{(n+3) !}{\left(n^2-4\right)(n+1) !}$
$\frac{(2 n+2) !}{(2 n) !}$
$\frac{1 !}{n !}=\frac{1 !}{4 !}-\frac{4}{5 !}$
$\frac{n}{6 !}=\frac{4}{8 !}+\frac{3}{6 !}$
$\frac{n}{8 !}=\frac{3}{6 !}+\frac{1 !}{4 !}$
$n =12, r =12$
$\begin{aligned} \therefore \quad \frac{ n !}{ r !( n - r ) !} & =\frac{12 !}{12 !(12-12) !}=\frac{12 !}{12 ! 0 !} \\ & =1 \quad \ldots[\because 0 !=1]\end{aligned}$
n = 8, r = 6
$\begin{aligned} \therefore \quad \frac{n !}{r !(n-r) !} & =\frac{8 !}{6 !(8-6) !}=\frac{8 \times 7 \times 6 !}{6 ! 2 !} \\ & =\frac{8 \times 7}{2 !}=\frac{8 \times 7}{1 \times 2}=28\end{aligned}$
$\frac{6 !-4 !}{4 !}$
$\frac{9 !}{3 ! 6 !}$
3! × 2!
10! – 6!