218 girls are to be made to stand in a row for a photograph. Among them three particular girls do not want to be together. In how many ways they can be arranged?
Explanation:
Required number of ways = Total ways - (Number of ways in which all three girls are together)
$G,G,G,G,G,\boxed{G_1,G_2,G_3}_X$
Let us group those three girls and name that group $X$
Now total number of objects $= 6$.
Number of ways of arranging Six objects $= 6!$
Now, the three girls can arrange themselves in $= 3!$ ways.
Total number of ways in which three girls are together $= 6! \times 3!$
Total ways in which three girls are not sitting together $= 8! - 6! \times 3!$
224 girls and 4 boys are to be made to stand in a row for a photograph so that girls and boys are standing alternatively. In how many ways they can be arranged?
Explanation:
Arranging the boys and girls alternatively can be done in following $2$ ways. A row can start with a boy or a girl.
$\quad\underline G \quad\underline B \quad\underline G\quad \underline B \quad\underline G\quad \underline B \quad\underline G \quad\underline B $
$ \quad\underline B \quad\underline G\quad \underline B \quad\underline G\quad \underline B \quad\underline G \quad\underline B \quad\underline G $
Number of ways of arranging $4$ boys in $4$ places $=4!$
Number of ways of arranging $4$ girls in $4$ places $=4!$
Total ways $= 2 \times 4! \times 4!$
234 girls and 4 boys are to be made to stand in a row for a photograph so that no two girls are together. In how many ways they can be arranged?
Explanation:
Arranging the boys in $4$ places provided $5$ places for girls shown in the below diagram.
$\quad\underline {\boxed1} \quad\underline B_1 \quad\underline {\boxed2}\quad \underline B_2 \quad\underline {\boxed3}\quad \underline B_3 \quad\underline {\boxed4} \quad\underline B_4 \quad\underline {\boxed5}$
Number of ways of arranging $4$ boys in $4$ places $=4!$
Number of ways of arranging $4$ girls in $5$ places $={}^5{P_4}$
Total arrangements $ = 4! \times {}^5{P_4}$
24Find the number of 3 digit numbers.
A1000 B899 C900 D901
Answer: C
Explanation:
$3$ digit numbers start from $100$ and end with $999$.
$\mathop {\boxed{\phantom{22222}}}\limits^{\boxed{1-9}} \qquad \mathop {\boxed{\phantom{22222}}}\limits^{\boxed{0-9}} \qquad \mathop {\boxed{\phantom{22222}}}\limits^{\boxed{0-9}}$
Number of ways of filling left most digit $=9$ ($\because $ $0$ should not be in the left most place)
Number of ways of filling tens digit $=10$
Number of ways of filling units digit $=10$
Total ways $=9 \times 10 \times 10 = 900$
25
Find the total numbers from 1 to 1000 (both inclusive)
A999 B1000 C1001 D1002
Answer: B
Explanation:
$ \mathop {\boxed{\phantom{22222}}}\limits^{\boxed{0-9}} \qquad \mathop {\boxed{\phantom{22222}}}\limits^{\boxed{0-9}} \qquad \mathop {\boxed{\phantom{22222}}}\limits^{\boxed{0-9}}$
By considering $0$'s in hundreds place and tens place we are including single digit and double digit numbers aswell.
Single digit numbers are the format = $\boxed{0} \qquad \boxed0 \qquad \boxed{k}$
Double digit numbers are the format = $\boxed{0} \qquad \boxed{l} \qquad \boxed{k}$
Number of ways of filling the left most digit $= 10$
Number of ways of filling the middle digit $= 10$
Number of ways of filling the units digit $= 10$
Total ways $= 10 \times 10 \times 10 = 1000$
Total numbers from $1$ to $999$ $ = 1000 - 1 = 999$ ($\because $ $000$ should be omitted)
Total numbers from $1$ to $1000$ $=999 + 1 = 1000$
26
Using the digits of the decimal system, how many numbers lying between 5000 and 6000 (both inclusive), can be formed if Repetition of digits is allowed.
A999 B1000 C1001 D1002
Answer: C
Explanation:
$\mathop {\boxed{5}} \qquad \mathop {\boxed{\phantom{22222}}}\limits^{\boxed{0-9}} \qquad \mathop {\boxed{\phantom{22222}}}\limits^{\boxed{0-9}} \qquad \mathop {\boxed{\phantom{22222}}}\limits^{\boxed{0-9}}$
We will consider the numbers from $5000$ to $5999$ so that we fix $5$ in the left most place.
Now we can change the digits in the remaining $3$ places in $10$ ways. i.e., $0$ to $9$
Total numbers start from $5$ $= 10 \times 10 \times 10$ $=1000$
Total numbers including number $6000$ also $= 1000 + 1 = 1001$
27
Using the digits of the decimal system, how many numbers lying between 5000 and 6000 (both inclusive), can be formed if Repetition of digits is not allowed.
A504 B900 C505 D1001
Answer: A
Explanation:
$\boxed{5} \qquad \boxed{\phantom{5}} \qquad \boxed{\phantom{5}} \qquad \boxed{\phantom{5}}$
We will consider the numbers from $5000$ to $5999$ so that we fix $5$ in the left most place.
Number of ways we can fill the hundreds digit $= 9$ ways ($\because $ $5$ is already used)
Number of ways we can fill the tens digit $= 8$ ways ($\because $ two digits are already used)
Number of ways we can fill the units digit $= 7$ ways ($\because $ three digits are already used)
Total number of ways $= 9 \times 8 \times 7 = 504$
28
Find the number of five digit number that can be formed by using the digits 2, 5, 6, 7, 9 without repetition so that The number is odd.
A72 B90 C120 D96
Answer: A
Explanation:
Given numbers are 2, 5, 6, 7, 9.
If a number has to be odd, it should end with 5, 7 or 9.
$ \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}}\quad\underline {\phantom{B}} \quad \underline {5/7/9}$
Number of ways of filling units digit = $3$
Number of ways of filling remaining digits = $4! = 24$
Total ways = $3 \times 24 = 72$
29
Find the number of five digit number that can be formed by using the digits 0, 2, 7, 8, 5, 6 without repetition so that the number is odd.
A199 B549 C192 D256
Answer: C
Explanation:
As the number formed is odd, they have to end with 7, 5. The number of ways in each case is shown below
$\quad \frac{{\phantom{0}}}{{ \times 0}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}}\quad\underline {\phantom{B}} \quad \underline {5/7}$
Number of ways of filling units digit $= 2$
Number of ways of filling left most place $= 4$
Number of ways of filling remaining digits $= 4! = 24$
Total ways $ = 2 \times 4 \times 24 = 192$
30
Find the number of six digit number that can be formed by using the digits 0, 2, 7, 8, 5, 6 without repetition so that The number is divisible by 5
A219 B259 C216 D256
Answer: C
Explanation:
Total digits = $6$
As the numbers formed are divisible by 5, each number has to end with either $0$ or $5$. a.Numebers ending with $0$
$\quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}}\quad\underline {\phantom{B}} \quad \underline 0$
Number of ways of filling $5$ digits in $5$ places = $5! = 120$ b.Numbers ending with $5$
$\quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}} \quad\underline {\phantom{B}}\quad\underline {\phantom{B}} \quad \underline 5$
Number of ways of filling left most place = $4$ ($\because$ $0$ should not be used, $5$ is already used)
Number of ways of filling remaining $4$ places $=4! = 24$
$\therefore $ Total numbers $= 120 + 96 = 216$
