11 Find the sum of all the five digit numbers formed by the digits 2, 4, 6, 8, 0 with repetition. A196110000 B176110000 C156110000 D136110000

Answer: D

Explanation:
Use the following Shortcut:
${{n^{n -1}}}$ $× (111... n\; times)$$ × (sum\;of\;the\;digits)$ $– {{n^{n - 2}}}$ $× (111...(n -1)\; times)$ $× (sum\; of\; the\; digits)$

Detailed Explanation:
Sum of digits at unit, ten’s hundred’s & thousand’s place $ = (0 + 2 + 4 + 6 + 8) × \bigl({5^4} - {5^3}\bigr)$ $ = 20 × 500 = 10000$
Sum of digits at ten thousand’s place $= (2 + 4 + 6 + 8) × {5^4}$ $= 20 × {5^4} = 12500$
Total Sum $= 12500 ×$ ${10^4}$ $× 10000 × {10^3}$$ + 10000 × {10^2}$ $+ 10000 × {10^1}$ $= 136110000$

12 I forgot my friend’s 7 digit telephone number, but I remembered that the first two digits of the number are either 25 or 53. Also the number is even and the digit 2 appears once. What is the maximum number of trials I have to make before I can contact my friend? A$56 \times {9^3}$ B$61 \times {9^3}$ C$64 \times {9^3}$ D$96 \times {9^3}$

Answer: B

Explanation:
There are two cases; dial 25 or 53 first. Case I
First two digits are $25$
$\quad\underline {2} \quad\underline {5} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}}\quad\underline {\phantom{B}} \quad\underline {4/6/8/0} $
Then total cases = $9 × 9 × 9 × 9 × 4$ $= {9^4} \times 4$ $= 36 \times {9^3}$

Case II
If we start with $53$ then either we can put 2 at unit place or at any of the remaining 4 places

Case II (i)If 2 is at units place.
$\quad\underline {5} \quad\underline {3} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}}\quad\underline {\phantom{B}} \quad\underline {2} $
Then, Total cases = $9 × 9 × 9 × 9 $ $= 9 \times {9^3}$

Case II (ii)If 2 is at any of the remaining 4 places.
$\quad\underline {5} \quad\underline {3} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}}\quad\underline {\phantom{B}} \quad\underline {4/6/8/0} $
$2$ can takes any of the four places.
Then, Total cases $= 4 \times {9^3} \times 4$ $ = 16 \times {9^3}$

13How many numbers greater than a billion but less than ${\rm{10}}^{{\rm{10}}} $ can be formed such that the sum of the digits is equal to 3? A54 B55 C72 D90

Answer: A

Explanation:
Billion is ${10^9}$. So it is a $10$ digit number, one followed by nine zeroes.
We have to find the number of ways such that the sum of digits is $3$.
Possible cases are,

Case I First digit is 1. Two of the remaining digits are ones, and remaining digits are zeroes.
$\;\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{1}\;\; \underbrace {\quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}}\quad\underline {\phantom{B}} \;{\phantom{B}}}_{9\;digits}\;\;$
Number of ways of arranging two one’s in 9 places = $^{{9}} {{C}}_{{2}} $ $= 36$ ways

Case II First digits is 1. One of the remaining digits $2$, and remaining digits are zeroes.
$\;\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{2}\;\; \underbrace {\quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}}\quad\underline {\phantom{B}} \;{\phantom{B}}}_{9\;digits}\;\;$
Number of ways of arranging one $2$ in $9$ places = $^{{9}} {{C}}_{{1}} $ $= 9$ ways

Case III First digits is $2$. One of the remaining digits $1$, and remaining digits are zeroes
Number of ways of arranging one $2$ in $9$ places = $^{{9}} {{C}}_{{1}} $ $= 9$ ways

Case IV First digit is $3$.
$\;\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{3}\;\; \underbrace {\quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}}\quad\underline {\phantom{B}} \;{\phantom{B}}}_{9\;digits}\;\;$
Only one way is possible if the starting digit is 3.
So total ways $= 36 + 9 + 9 + 1 = 55$ ways

14How many six digit numbers are possible by using the digits 1, 2, 3, 4, 5, 6, 7 without repetition such that they are divisible by 12? A$20 \times 4!$ B$22 \times 4!$ C$24 \times 4!$ D$26 \times 4!$

Answer: C

Explanation:
To be divisible by 12, a number should be divisible by both $3$ & $4$

Sum of all the available digits is 28.
We have to pick any six digits such that their sum is divisible by 3.
Possible sums less than 28 are 27, 24, 21 .
For the sum to be $21$, leave 7. For the sum to be 24, leave 4. For the sum to be 28, leave 1.

Case (i) 1, 2, 3, 4, 5, 6
This number will be divisible by 4 if the last two digits are
$12, 16, 24, 32, 36, 52, 56, 64$
For all the above 8 cases, the remaining 4 digits can be arranged in 4! Ways.
Number of ways $= 4! \times 8.$

Case (ii) 1, 2, 3, 5, 6, 7
Last two digits can be $12, 32, 52, 72, 16, 36, 56, 76$
Number of ways = $4! \times 8$

Case (iii) 2, 3, 4, 5, 6, 7.
Last two digits can be $24, 32, 36, 52, 56, 64, 72, 76$
Number of ways = $4! \times 8$

Total number of ways $= 24 \times 4!$

15 All the possible seven digit numbers are formed by using the digits 1, 2, ........., 7 without repetition. These numbers are then arranged in ascending order. What will be the ${\rm{2884}}^{{\rm{th}}} $ number in the given order? A5123476 B5123647 C5123674 D5123467

Answer: C

Explanation:
Total Possible seven digits numbers $= 7!$ $= 5040$
Each of these 7 digits will be the starting digits for $\displaystyle\frac{{5040}}{7}$ = 720nos
So 1, 2, 3, 4 will be the starting digits for first 2880 numbers when the numbers are arranged in ascending order.
2881 number $ \to $ 5123467
2882 number $ \to $ 5123476
2883 number $ \to $ 5123647
2884 number $ \to $ 5123674
$\therefore $ ${\rm{2884}}^{{\rm{th}}} $ number = 5123674.

16 In how many ways can a mixed doubles tennis game be arranged from eight married couples, if no husband and wife play in the same game? A3199960 B4199960 C5199960 D6199960

Answer: C

Explanation:
There are eight married couples or eight males and eight females. First we can select 2 males out of 8 males in $^8 C_2 $ ways. Now according to the condition given in the question wives of these 2 males can not be in the same game so we can select any 2 females from the rest 6 females in $^6 C_2 $ways. Than taking these four persons we can arrange two different matches by changing the partners.
Hence total no.of matches = $^{\rm{8}} {\rm{C}}_{\rm{2}} {\rm{ \times }}^{\rm{6}} {\rm{C}}_{\rm{2}} $ x 2 = 28 x 15 x 2 = 840

17 Sixteen persons are to be seated along two sides of a rectangular table with eight chairs on either side. Of these sixteen, Amit, Jaya, Rishi and Neetu with to sit on one side of the table, while Paresh and Sadashiv with to sit on the other, in how many ways can the arrangement be done? A${}^8{C_4} \times {}^8{C_2} \times 10!$ B${}^8{P_4} \times {}^8{P_2} \times 10! $ C${}^8{C_4} \times {}^8{C_2} \times 10! \times 2$ D${}^8{P_4} \times {}^8{P_2} \times 10! \times 2$

Answer: D

Explanation:
$\quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad {(1)}$
$\quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad {(2)}$
As shown in the figure there is a rectangular table and sixteen persons are to be seated on the sixteen chairs along two opposite sides of the table, eight chairs on either side. Among these sixteen persons four want to sit on the same side so we can select that side in two ways and as given in the question two other persons want to sit on the other side of the table.

Now after selecting the side for four persons we can arrange them on eight chairs in $^{\rm{8}} {\rm{P}}_{\rm{4}} $ ways, similarly other two persons on the opposite side can be arranged on eight chairs in $^{\rm{8}} {\rm{P}}_{\rm{2}} $ ways. Rest 10 persons can be arranged on remaining 10 chairs in 10! Ways.
So the total no.of ways to arrange these 16 persons considering all the conditions = ${}^8{P_4} \times {}^8{P_2} \times 10! \times 2$

18 Six people A, B, C, D, E and F are to be seated at a circular table. In how many ways can this be done if A must always have either B or C on his immediate left and B must always have either C or D on his immediate left? A24 B18 C48 D96

Answer: B

Explanation: Case (i)
If A has B on his left then B can have either C or D on his left & then remaining 3 persons can arrange themselves in 3! = 6 ways.
So total ways are 2 x 6 = 12 ways.

Case (ii)
If A has C on his left then B & D will sit such that D is always on left of B & this can be done in 3! = 6.