Exercise 1.1
1. Solution:
Total number of air
flights (n1) = 5
Total number of buses (n2)
= 15
As from the addition
rule
The number of ways to
travel from Bhairahawa to Kathmandu = 5 + 15 = 20
Hence, there are 20 ways
to travel.
2. Solution:
The
number of girls and boys are 25 and 20 respectively. If a boy and a girl are to
be chosen for debate competition, then
The number of ways of
selection would be 25×20 = 500 ways.
3. Solution:
If there are 5 routes
from station A to station B and 4 routes from station B to C then,
The number of possible
routes from A to B is 5 and from B to C is 4.
a. Here,
∴The number of possible
routes from A to C is 5×4 = 20 routes.
b. Here,
The number of possible
routes from A to C is 20 and so as to return from C to A there is also 20
routes (i.e. 4×5).
Hence, the number of
required ways = 20×20 = 400 ways
c. Here,
The number of routes to
travel from A to C is 20. If the same route is not used more than once, then
The number of ways to
travel and return back is 20×12 = 240 ways
4. Solution:
The number of digits = 6
So, hundred place can be
arranged in 6 ways
Tens place can be
arranged in 5 ways
Units place can be
arranged in 4 ways
∴Required numbers = 6×5×4
= 120
Next, if these numbers
formed must be even, the digit in the units place can be arranged in 3 ways
Ten's place can be
arranged in 5 ways
Hundred place can be
arranged in 4 ways
∴Required number's place
= 3×5×4 = 60 ways
5. Solution:
The number of digits = 6
a. As we know, units place
can never be filled by zero, so units place can be filled by 5 ways
Tens place can be filled
by 5 ways
Hundred place can be
filled by 4 ways
Thousand place can be
filled by 3 ways
∴The required numbers of
4 digit when repetition is not allowed = 5×5×4×3 = 300
b.
If the repetition is allowed, the unit place can be arranged/filled by 5 ways
and then after all remaining places can be filled by 6 ways.
∴The required numbers of
4 digit when repetition is allow = 5×6×6×6 = 1080 ways
6. Solution:
The given digits are 0,
1, 2, and 3
If the digits may
repeat: then
For 1 digits: For the
units place, number of way = 4
For the ten's place,
number of ways = 3
∴ Number of ways = 4
× 3 = 12
For 1 digit: The number
of ways = 3
So, total number of ways
= 12 + 3 = 15
If the digits may not
repeat:
For 1 digit: Number of
ways = 3
For two digits = Number
of ways in tens place = 3
Number of ways in ones
place = 3
∴ Number of ways =
3×3 = 9
So, total number of ways
= 3 + 9 = 12
7. Solution:
The number of digits = 5
The number must lies
between 2000 and 3000 and so each number should be started with 2.
As the formed number
should be even each number must be ended with 0, or 2 but here digits can be
used only once.
So, units place can be
filled by 1 ways
Tens place can be filled
by 4 ways
Hundred place can be
filled by 3 ways
Thousand place can be
filled by 1 ways
∴Required number of
digits = 1×4×3×1 = 12
8. Solution:
Here, the numbers are 0,
1, 2, 3, 4, 5, 6, 7, 8, 9 and the telephone number starts with 562
i. If repetition is not allowed: Number of ways for
remaining 3 places = 7×6×5 = 210
ii. If repetition is allowed: Number of ways for
remaining 3 places = 10×10×10 = 1000
9. Solution:
The total digit is 10
and number of choice for unit digit is 9.
So the number digits required
= 8×9×9 = 648
Now, if there is only
one zero given that the repetition allowed,
Then, The unit place
filled by 9 ways
Tens place can be filled
by 10 ways
Hundred place can be
filled by 9 ways
∴ Number of total
digits = 9×10×9 = 810
Hence, the required
number = 810 – 648 = 162
Exercise 1.2
1. Solution
Given, \(\frac{{\left(
{{\rm{n\;}} + {\rm{\;}}1} \right)!}}{{\left( {n - 1} \right)!}} = {\rm{\;}}12\)
Or,\(\frac{{\left(
{{\rm{n\;}} + {\rm{\;}}1} \right){\rm{n}}\left( {{\rm{n}} - 1} \right)!}}{{\left(
{n - 1} \right)!}} = {\rm{\;}}12\)
Or, n (n+ 1) = 12
Or, n (n + 1) = 12
Or, n2+ n –
12 = 0
Or, n2+ 4n –
3n – 12 = 0
Or, (n + 4) (n – 3) = 0
Either n = –4
Or, n = 3
Since, n ≠ –4, so n = 3
2. Solution:
Given, P (5, r) = 5
r = 1 [∴ if P (n, r) = n, then r = 1)
3. Solution:
The no. of digits (n) =
6
a. Units place can be
filled only by 5 digits but the remaining 3 places can be filled 6 digits as
the repetition is allowed.
∴The required no. of 4
digits = 5×6×6×6= 1080
b. Unit first place can
be filled by 5 digit as the repletion not allowed.
2nd first
place can be filled by 5 digit
3rd first
place can be filled by 4 digit
4th first
place can be filled by 3 digit
∴The required no. of 4
digit = 5×5×4×3= 300
4. Solution:
There are 4 boys and 3
girls be seated in a row containing 7 seats.
∴Required arrangement is
P (7, 7) = \(\frac{{7!}}{{\left( {7 - 7} \right)!}} = \frac{{7!}}{{0!}} =
5040\)
Again,
If they seat
alternatively, then 4 boys can set in 4! Ways and 3 girls can seat in 3! Ways.
∴Required arrangement is
= 4! × 3!
= 24×6 = 144 ways
5. Solution:
The total no. of digits
= 10
The first digit can be
chosen from only 1 to 9 so there is only 9 choices for first digit.
The remaining 5 digits
can be chosen from remaining 9 digits in p(95) ways
i.e. \(\frac{{9!}}{{\left(
{9 - 5} \right)!}} = 15120Ways\)
∴The total numbers of 6
digits is 9×15120 way = 136080
Next: For the divisible
by 10. Last digit must be zero, so the last digit can be chosen from 0, so
there is 1 choice for last digit. The remaining 5 digits can be chosen from 9
digits in P (9, 5) way
i.e.\(\frac{{9!}}{{\left(
{9 - 5} \right)!}} = 15120Ways\)
6. Solution:
The numbers given in the
question is 1, 2, 3, 4, and 5
For one digit: No. of
ways for even = 2
For two digits: No. of
ways for one’s place = 2
Number of ways for ten's
place = 4
∴ Total no. of ways
2 + 2×4 = 10
7. Solution:
In a bracelet, beads are
arrangement in circular form and the anticlockwise and clockwise arrangements
are not different.
Here the total number of
beds n = 9
They can be arranged in
(n – 1)! Ways = ½ × 8! Ways = 20160
8. Solution:
The no. lying between
100 and 1000 is of 3 digit. In which at unit place can be chosen only from 5
digit and hundred place can only be chosen from 5 digit whereas remaining tens
place can be chosen from remaining 4 digit.
∴The no. formed between
100 and 1000 = \(\frac{{5!}}{{\left( {5 - 1}
\right)!}} \times {\rm{\;}}\frac{{4!}}{{\left( {4 - 1} \right)!}} \times
\;\frac{{5!}}{{\left( {5 - 1} \right)!}}\)
\( = \;\frac{{5 \times 4!}}{{4!}} \times
\frac{{4 \times 3!}}{{3!}} \times \frac{{5 \times 4!}}{{4!}} = 100\)
9. Solution:
Each post cards can be
posted in 4 ways
Hence, required number
of ways = nr= 55= 1024
10. Solution:
a. PERMUTATION
Here, Total no. of
letters (n) = 11
No. of letter 'T' (p) =
2
∴Total number of way of
arrangement = \(\frac{{n!}}{{p!}} =
\frac{{11!}}{{2!}}\)
b. INTERMEDIATE
Here, the total number
of letters (n) = 12
No. of letter 'I' (p) =
2
No. of letter 'T' (q) =
2
No. of letter 'E' (r) =
3
∴The total no. of arrangement
= \(\frac{{n!}}{{p!q!r!}} = {\rm{\;}}\frac{{12!}}{{2!2!3!}}\)
c. EXAMINATION
Here, the total number
of letter (n) = 11
No. of letters 'A' (p) =
2
No. of letters 'I' (q) =
2
No. of letters 'N' (r) =
2
∴The total no. of
arrangement = \(\frac{{n!}}{{p!q!r!}} =
\frac{{11!}}{{2!2!2!}}\)
d. CIVILIZATION
The total no. of letters
(n) = 12
No. of letter 'I' (p) =
4
∴Total number of way of
arrangement = \(\frac{{n!}}{{p!}} =
\frac{{12!}}{{4!}}\)
12. Solution:
In UNIVERSITY'
The no. of letters (n) =
10
No. of letter 'I' (p) =
2
∴The total no. of
arrangement = \(\frac{{n!}}{{p!}} =
\frac{{10!}}{{2!}} = 1814400\)
Since the arrangement
begin with U there is only. Nine letters to arrange. So, the nine letters can
be arranged in \(\frac{{n!}}{{p!}} = \frac{{9!}}{{2!}}
= 181440\)
∴Required no. of
arrangement = 1×181440 = 181440
Next: The total no. of
ways in which the arrangement begin with U but do not end with
13. Solution:
Total no. of countries
(n) = 8
If they sit in round
table then they form a circle, so its arrangement
= (n – 1)! = 7! =5040
If Nepali and Indian
always sit together, then we take it as one. Then the total no. will be 6.
So, the arrangement
(n–r+1)! = (7–2+1)! = 6! = 720
If they sit together,
then they also can interchange there seat between themselves in 2 ways.
Hence, the required no.
of arrangement = 2×720 = 1440
14. Solution:
If there is 3 candidate for the president then election can be
turned in 3 ways. Similarly for 5 secretary and 2 pressure the election can be
turned in 5 ways and 2 ways. Hence, the required no. ways to conduct election =
3! × 5! ×2! = 1440
15. Solution:
Total no. of digit = 4
The person can try his
password in P (4, 4) ways = \(\frac{{4!}}{{\left( {4 - 4}
\right)!}} = 24\)
16. Solution:
Since 6 persons are to be arranged in row with 6 seat, so that the
girls and boys are in alternate, so girl are to be arranged in odd seats and
boy in even seats.
∴The total no. of
arrangement = 6! = 720
Here, the no. of
arrangement of boy restricted to occupy even seats is P (3, 3) = \(\frac{{3!}}{{\left(
{3 - 3} \right)!}} = 6\)
The 3 boys can occupied
seats in 3! By interchanging their seats.
Hence, the required no.
of arrangement = 3! = 36
17. Solution:
In 'EQUATION'
The no. of total letters
(n) = 8
∴The total no. of
arrangement = 8! = 40320
Next, The no. of vowels
= 5
When we take all vowels
as one then there will be total letters left = 4. Also the vowel letters be
arranged themselves in 5 ways.
∴The required no. of arrangement = 4! × 5! = 2880
Exercise 1.3
