Permutations and Combinations Class 11

Ex6.1, 1 teackoo.com
How many 3-digit numbers can be formed from the digits 1, 2, 3, 4
and 5 assuming that
(i) repetition of the digits is allowed?
3 digit number :
youd
15t Qnd gid
digit digit digit
Position of Number of
digit different ways
1% 5
2nd 5
3M 5
Number of 3 digit numbers with repetition= 5 x 5 x 5
=125

How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that
(ii) repetition of the digits is not allowed?

Ex 6.1, 2 teachoo.com
How many 3 digit even numbers can be formed from the digits
1, 2, 3, 4, 5, 6 if the digits can be repeated?
Let the 3 digit even number be
Rough
TT TT TT 158 is even as 8 is even
Hundreds Tens Units 296 is even as 6 is even
Place Place Place
Only 3 numbers are possible at units place (2 , 4 & 6) as we need
even number.
But at tens & hundreds place all 6 are possible
Number of
Different ways
Number of 3 digit even numbers Units 3
=3x6x6=108 Tens 6
Hundreds 6

Ex 6.1, 3 teachoo.com
How many 4-letter code can be formed using the first 10 letters of th
English alphabet, if no letter can be repeated?
No. of letters = 10
Code is of the form
yr V4
4st 2nd grd qth
digit digit digit digit
7 Number of
Letter Position Different ways
1° position 10
2° position 9
3" position 8
4" position 7
Total number of possible 4 - letter code = 10 x 9x 8x 7= 5040
Hence 5040 ways 4 letter code can be formed

Ex 6.1, 4 teachoo.com
How many 5—digit telephone numbers can be constructed using the
digits 0 to 9 if each number starts with 67 and no digit appears
more than once?
The telephone number is of 5 digits and
first 2 numbers are 6 and 7
Hence number is of the form Number of
different ways
st fj:
6 7 —_ _ 1 fix 1
4; 4 4 + 4 2" fix 1
qst pnd 3rd qth 5th
digit digit digit digit digit 3m 8
qh 7
sth 6
Digits between 0 to 9 = 0, 1, 2, 3, 4, 5, 6, 7, 8,9

teachoo.co
Ex 6.1, 5 aehOo.com
A coin is tossed 3 times and the outcomes are recorded.
How many possible outcomes are there?
A coin is tossed 3 times & the outcomes are recorded
outcomes outcomes

First Head, Tail 2

Second Head, Tail 2

Third Head, Tail 2
Total number of possible outcomesis 2 x 2 x 2

=8

teachoo.com
Ex 6.1, 6
Given 5 flags of different colours, how many different signals
can be generated if each signal requires the use of 2 flags, one
below the other?
A signal can have only 2 flags
[| — First Flag
[| — Second Flag
Flag Position |Number of
Different ways
First 5
Second 4
The required number of signals =5 x 4
=20

Ex 6.2,1 teachoo.com
Evaluate
(i) 8!
8!
=1x2x3x4x5x6x7x8
= 40320

Evaluate
(ii) 4! 3!

teachoo.com

Ex 6,2, 2
Is 31+ 41 = 71?

L.H.S R.H.S
31+4! 7!
=3x2x1+4x3x2xi1 =1x2x3x4x5x6x7
=6+24 = 5040
= 30
Hence 3! + 4! #7!

teachoo.com
Ex 6.2, 3
8!
Compute ———
61x 2!
8!
61x 2!

_ 8x 7x 6!

“61x (2x 1)

_8Xx7

~ 2

=4x7

=28

Ex 6.2, 4 teachoo.com
1 1 x.
If [+5 25 find x.
1 1 x
S4+os—
6! 7!” 8B!
1 1
8! 8!
etn
8x 7x 61 8x 7!_
6! +X
8x7+8=x
56+8=x
64=x
x = 64

Ex 6.2, 5 teachoo.com
Evaluate 1 , when
@-r)!
(i) n=6,r=2
n!
(n-r)!
Puttingn=6&r=2
6!
~ (6-2)!
6!
“Al
_ 6x 5x4!
"Al
=6x5
= 30

Evaluate n!/(n − r)! , when
(ii) n = 9, r = 5

Ex 6.3, 1 (Method 1) teachoo.com
How many 3-digit numbers can be formed by using the digits
1 to 9 if no digit is repeated?
Let the 3-digit number be
1 1 T Number of
r Different
Hundred Tens Units _
Digits
place _ place place
Units 9
Tens 8
Hundred 7
Number of 3 digit number which can be formed =9x 8x 7
= 504 ways

Ex 6.3, 2 (Method 1) teachoo.com
How many 4-digit numbers are there with no digit repeated?
We need to make 4 digit numbers using digits 0,1,2,3,4,5,6,7,8,9
But, these include numbers starting with '0' like 0645, 0932, ...etc

which are actually 3 digit numbers
Required numbers

=" Total " 4" digit numbers "
-4 digit number which have 0 in the beginning

Ex 6.3, 3 (Method 1) teachoo.com
How many 3-digit even numbers can be made using the digits 1, 2,
3, 4, 6, 7, if no digit is repeated?
We need to find 3 digit even number using 1, 2, 3, 4, 6, 7,
Hence units place can have either 2, 4 or 6
296 is even as 6 is even
Number of even numbers if 2 is at units place
2
Hence these are 5 more digits left (1, 3, 4, 6, 7) for
Hencen=5
which we need to fill 2 place
andr=2
Number of 3 digit even number with 2 at unit place ="P, = °P,

Ex 6.3, 4 (Method 1} teachoo.com
Find the number of 4-digit numbers that can be formed using the
digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be
even?
Let the 4 digit number be
Total number of digits (1, 2, 3, 4,5) =5

So,n=5
We need to make 4-digit number, so we take 4 digits at a time,

So,r=4
Number of 4-digit numbers = "P,

=5P,

Ex 6.3, 5 (Method 1) teachoo.com
From a committee of 8 persons, in how many ways can we choose
a chairman and a vice chairman assuming one person cannot hold
more than one position?
We need to choose a chairman & a vice-chairman out of 8 person
Here n = Number of people = 8
& r = Number of people to be chosen = 2
Number of ways choosing a vice-chairman = ®P,
8!
~ (8-2)!
8!
*6l
_8X7X6l
“6!
=8 x7
= 56 ways

teachoo.com
Ex 6.3, 6
Find nif °~1P,:°P,=1:9.
Lets first calculate
"1p. and "P, separately ' @—n!
n-1p, Py
(n)!
_ (m-1)! =o
~ (n-1 -3)! (n-4)!
_ n(n—1)(n—2)(n—3)(n—-4)!
=a} - (na)!
(ma = n(n —1)(n—2)(n— 3)
_ (n-1)(n- 2)(n- 3)(n- 4)!
~ (n—4)!
=(n—-1)(n—- 2)(n—-3)

Ex 6.3, 7 teachoo.com
Find r if
(i) 53P,=2 °P,_,
np =—™
Calculating °P, & °P,_, "Mm!
5P, §Pi_4
5! _ 6!
“(S=r)! ~ (6 —(r-1))!
6!
~ (6-r+1)!
6!
“(7=-r)!
Given,
°P, = 2°P 4

Find r if
(ii) 5Pr = 6Pr – 1

Ex 6.3, 8 teachoo.com
How many words, with or without meaning, can be formed using all
the letters of the word EQUATION, using each letter Exactly once?
Number of letters in word ‘EQUATION = 8
n=8
If all letters of the word used at a time
r=8
Different numbers formed = "P,
= 8p,
3!
~ (8-8)!
3!
*o

Ex 6.3, 9 teachoo.com
How many words, with or without meaning can be made from
the letters of the word MONDAY, assuming that no letter is
repeated, if

(i) 4 letters are used at a time,
Total number of alphabets in MONDAY = 6

Hence n=6

If 4 letters are used at a time,

r=4
Number of different words ="P,
=P,
6!
~ (6-4)!

Ex 6.3, 10 teachoo.com
In how many of the distinct permutations of the letters in
MISSISSIPPI do the four I’s not come together?
Total number of permutation of 41 not coming together
= Total permutation

— Total permutation of | coming together
Total Permutations
In MISSISSIPPI
there are 41, 4S, 2P and 1M
Since letters are repeating,
we will use the formula = a

p,!p,!p,!

Ex 6.3, 11 teachoo.com
In how many ways can the letters of the word PERMUTATIONS
be arranged if the
(i) words start with P and end with S
Let first position be P & last position be $ (both are fixed)
10 letters in which 2T

Since letters are repeating
Hence we use this formula —-—

P,!p,'p,!
Total number of letters = n = 10

Ex 6.4, 1 teachoo.com
If "Cy = "C,, find °C, .
Given "C, ="C,
if"C,="C,
then eitherr=porr=n—p
Thus, "C, ="C,
8=n-2
8+2=n
10=n
n=10

Ex 6.4, 2 teachoo.com
Determine n if
(i) 2°C3:°C3=12:1
Let first calculate 2"C, and"C, separately
nr e ni
C= ria—r)!
2nC, "C;
2n! =”
= 3(2n—3)! 31(n - 3)!
(n(n -1)(n-2) (2-3)!
_ (nj Qn -1)(2n—2)(2n—3)! =
=" ~"3x2 x 1)(2n—3)! 3X2 x 1)(n-3)!
n(n-1)(n-2)
_ 2n(2n—1)(2n-2) =——
~ 6
Given
ne, :"C;=12:1
menor? aoe =12:1

Determine n if
(ii) 2nC3 : nC3 = 11 : 1

Ex 6.4, 3 teachoo.com
How many chords can be drawn through 21 points on a circle?
. A
Number of points = 21 B

Hencen=21
A chord connects circle at 2 points
Hence, r =2
Number of chord from 21 points = "C,
- 21¢,
_ 21!
~ 21(21-2)
_ 21!
~ 2119!
_ 21x 20x 19!
~ (2x1) x 19!
_ 21x 20
~ 2
= 210 chords.

Ex 6.4, 4 teachoo.com
In how many ways can a team of 3 boys and 3 girls be selected from
5 boys and 4 girls?
Selecting 3 Boys Selecting 3 girls
We have to select Similarly we have to select
3 boys from 5 boys 3 girls from 4 girls
Here,n=5&r=3 Heren=4& r=3
Number of ways ="C, Number of ways = "C,
=5C, = ‘C,
= 4
31(5 3)! ~ 314-3)!
= 2) _ 4x3!
312! ~ 31x14!
=10 =4

Ex 6.4, 5 teachoo.com
Find the number of ways of selecting 9 balls from 6 red balls, S white
balls and 5 blue balls if each selection consists of 3 balls of each
colour.
3 Red balls 3 White balls 3 Blue balls
We have to select 3 | We have to select 3 We have to select 3
red balls from 6 red | white balls from 5 blue balls from 5 blue
balls white balls balls
Here, n=6 &r =3 Here,n=5&r=3 Here,n=5&r=3
No of ways of No of ways of No of ways of
selecting 3 red balls | selecting 3 white balls | selecting 3 white balls

= 6C, =5C; =5C,

6! 5! 5!
~ 36 =pi= 20 ~ 345-3) 10 ~ 35-3)! 10

Ex 6.4, 6 teachoo.com
Determine the number of 5 card combinations out of a deck of 52
cards if there is exactly one ace in each combination.
Number tobe | Number of ways
Total number
chosen to choose
Ace 4 1 4c,
Other 48 4 48C
‘4
52 5
Total number of ways = 4C, x “°C,
4 48!
= ——_ x ——_
1(4-1)! 41(48 —4)!
4 48!
=— x —
13!“ 4l4a!
_ 48!
~ 31x 44!

Ex 6.4, 7 teachoo.com
In how many ways can one select a cricket team of eleven
from 17 players in which only 5 players can bowl.
If each cricket team of 11 must include exactly 4 bowlers?
Number to be Number of
Total number
chosen ways to choose
Bowler 5 4 5Cy
Other 12 7 LC,
17 11
Total number ways = °C, x 1C,
5! 12!
=— x —
all 715]
_ 12!
~ atx 7!
_12X 11x 10x 9x 8x 7!
"(4x 3x 2x 1)x 7!
= 3960

Ao.
Ex 6.4, 8 teachoo.com
A bag contains 5 black and 6 red balls. Determine the number of
ways in which 2 black and 3 red balls can be selected.
We have to select 2 black balls We have to select 3 red balls
from 5 black balls from 6 red balls
Here, n=5 Heren=6
& r=2 & r=3
No. of ways selecting 2 black No of ways selecting 3 red balls
from 5 black = °C, from 5 red balls = °C,
__ 3} 6!
~ 215-2)! ~ 36 —3)!
_ 3} 6!
~ 2131 ~ 318
_ 5x4x3! _6xX5xX4x 3)
"2x3! ~3x2x1Xx3!
_ 5x4 _6x5x4
~ 2 “3x2
=10 =20

Ex 6.4, 9 teachoo.com
In how many ways can a student choose a programme of 5
courses if 9 courses are available and 2 specific courses are
compulsory for every student?
Total Number to
number | be chosen
Compulsory Courses 2 2
Other courses 7 3
9 5
Hence, the student has to select 3 courses out of remaining
7 courses
Here, n=7
& r=3
No of ways selecting 3 course out of 7 course = "C,
= "Cy
7
~ 31(7 -3)!

Example 1 teachoo.com
Find the number of 4 letter words, with or without meaning,
which can be formed out of the letters of the word ROSE, where
the repetition of the letters is not allowed .
First Second Third Fourth
Place Place Place Place
Number of letters which can be filled in different places are
Pl Number of
ace letters filled
1st 4
and 3
3rd 2
4m 1
Number of ways in which 4 places can be filled =4x 3x 2x 1

Example 2 teachoo.com
Given 4 flags of different colours, how many different signals
can be generated, if a signal requires the use of 2 flags one
below the other?
Let the signal be of the form
[ ] — First Flag
| — Second Flag
Flag Position |Number of
Different ways
First 4
Second 3
The required number of signals
=4x 3=12

Example 3 teachoo.com
How many 2 digit even numbers can be formed from the
digits 1, 2, 3, 4, 5 if the digits can be repeated?
Let the 2 digit even number be
J J 18 is even as 8 is even
Tens Units 26 is even as 6 is even
Place Place
Only 2 numbers are possible at units place (2 and 4) as we
need even number. But at tens place all 5 are possible
Number of
Place
Different ways
Units 2
Tens 5
Number of 2 digit even numbers = 2x 5 =10

Example 4 teachoo.com
Find the number of different signals that can be generated by
arranging at least 2 flags in order (one below the other) ona
vertical staff, if five different flags are available?
A signal can have at least 2 flags
i.e. flag can have 2 flags or 3 flags or 4 flags or 5 flags
We need to calculate separately for each and then add
For 2 flags
[ ] — First Flag Flag Position Number of
Different ways
| —+ Second Flag 5
First 5
Second 4
Required number of signals = 5 x 4=20

Example 5 teachoo.com
Evaluate
(i) 5!
Blalx2x3x4x5
=120
(ii) 7!
7W!=1x2x3x4x5x6x7
= 5040
(iii) 71-5!
7!-5!
= 5040 - 120
= 4920.

teachoo.com
Example 6
Compute
.7!
OF
7
5!
_7x6xS5!
“51
=7x6
=42

Example 6
Compute
(ii) 12!/((10!)(2!))

teachoo.com
Example 7
n!
Evaluate ——— when n =5,r =2.
ri(n —r}!
nl
rin-r)!
Puttingn=5,r=2
5)
21(5 -2)!
5]
~ 21x 3!
_5 x4 x3!
"21x 31
_5 x4
"2x1
20
"2
= 10

Example 8 teachoo.com
1 1 x .
fata Zap. find x.
1 1 x
+= —
8! 91 10)
1 1
10! 10) _
‘ato %
10x 9x 81 10x 9!
—,— +. = x
8! 9)
10x9+10=x
90+ 10=x
100 =x
Hence, x = 100

Example 9 teackoo.com
Find the number of permutations of the letters of the word
ALLAHABAD.
In ALLAHABAD
There are 4A, 2L, 1H, 1B &1D
Since alphabets are repeating we will us this formula an
Total number of alphabets = 9
Heren=9,
There are 4A’s, 2l’s
hence taking p, =4 & p, =2
Hence total number of permutation = ~
91
* 72

Example 10(Method 1) teachoo.com
How many 4-digit numbers can be formed by using the digits 1
to 9 if repetition of digits is not allowed?
n= Numbers from 1 to9 =9
r=4
(as we gave to take 4 digits at a time)
Required 4 digit number
=P,
91
~ (9-4)!
91
*3
_9X8X7X6X 5!
~ 5!
= 3024

Example 11 (Method 1) teachoo.com
How many numbers lying between 100 and 1000 can be formed
with the digits 0, 1, 2, 3, 4, 5, if the repetition of the digits is not
allowed?
Number between 100 and 1000 will be 3 digit number
3 digit number will be formed from the digits 0, 1, 2, 3, 4, 5
But, these include numbers starting with ‘0’ like 051, 012, ...etc
which are actually 2 digit numbers
Required numbers
= Total 3 digit numbers
— 3 digit number which have 0 in the beginning

Example 12 teachoo.com
Find the value of n such that
(i) "P,=42 "P,,n >4
np ___7!
Given
np, = 42 "P,
Calculating "P, Calculating 42"P,
1 n = nl
"p, = a 42"P, = 420
n(n -1(n—2)(n—3)(n— a(n 5)! = Hate n= en 9)
= rr | (n - 3)!
=n(n—1)(n—2)(n—3)(n—4) = 4an(n— 1}{n— 2)

Example 12
Find the value of n such that
(ii) "nP4" /"n−1P4" = 5/3 , n > 4

Example 13 teachoo.com
Findr, if 54P,=6°P,,
!
Now it is given that
54P, = 6°P
Al 5}
Al 3!
3* Gon =8* Gee
4! 3!
5x Gun 6x @onI
(6-r)! _ 6x5!
(4-r)! 5x4!
(6-r)\(S-r)(4—-r)! _ 6x5!
(-r)} "5x4!
6x5!

Example 14 teachoo.com
Find the number of different 8-letter arrangements that can be
made from the letters of the word DAUGHTER so that

{i) all vowels occur together

Total number of letter in DAUGHTER = 8

Vowels in DAUGHTER=A, U &E (Vowels are a, e, i, 0, u)
Since all vowels occur together,

Assume as single object.

So, our letters become [D | [R |

Example 15 teachoo.com
In how many ways can 4 red, 3 yellow and 2 green discs be
arranged in a row if the discs of the same colour are
indistinguishable ?
Total number of discs = 4 red + 3yellow + 2 green
n =9.
Since color are repeating so we use this formula a
p,!p,!p,!
Heren=9
Out of 9 discs,
4 red, 3 yellow and 2 green
& p, = 4, P2 = 3, P3= 2
n!
The number of arrangement = ———
P,'P,'p3!
91
~ qy3t2i

Example 16 teachoo.com
Find the number of arrangements of the letters of the word
INDEPENDENCE. In how many of these arrangements,
Finding total number of arrangements
In word INDEPENDENCE
There are 3N, 4E, & 2D, 11, 1P & 1C
Since letters are repeating, so we use this formula

nl!

p,!p,'p,!

Total letters = 12
So,n=12

Example 17 (Method 1) teackoo.com
If "Co = Cg, find "C,7 .
"Cy = NC, h _ nl
ni _ ni
9(n-9)! (n—8)!8!
ni(n — 8)! a
(1-9)! 8!
(n - 8)! _ a
(1-9)! "8!
(n-8)(—-8—-1)!_9x8!
(n-9)! ~ 8!
(n-8)(m—-9))_
(n-9)) 9
n-8=9
n=17

Example, 18 teachoo.com
A committee of 3 persons is to be constituted from a group of 2
men and 3 women. In how many ways can this be done? How
many of these committees would consist of 1 man and 2 women?
We have to select a committee of 5 different person taken 3 at a
time
Here,n=5
r=3

The required number of ways = "C,

=5C;

5]
~ 315 -3)!
5!

~ 3121

_ 3x4x 3!

~ 3Ix(2x D

=5x2

= 10 ways

Example 19 teachoo.com
What is the number of ways of choosing 4 cards from a pack of
52 playing cards? In how many of these
(i) four cards are of the same suit,
There are four suits
i.e. Diamond, Spade, Heart, Club
& 13 cards of each suit
a=
be chosen | of ways
Diamond 13 4 BC,

Spade 13 4 8¢,

Heart 13 4 BC,

Club 13 4 BE,

Example 20 teachoo.com
How many words, with or without meaning, each of 3 vowels and 2
consonants can be formed from the letters of the word INVOLUTE ?
oN
Vowels Consonants
L0,UE NVLT
Total Number to be | Number of ways
number chosen to choose
Vowels 4 3 4G,
Consonants 4 2 4c,
Thus,
Number ways of selecting 3 vowels & 2 consonants
=4C, x 4C,

Example 21 teachoo.com
A group consists of 4 girls and 7 boys. In how many ways can a team
of 5 members be selected if the team has
(i) no girl ?
Total | Number to | Number of ways
number | be chosen to choose
Girls 4 0 4c,
Boys 7 5 om
Total number of ways = 4C, x ’C,
4! 7
~ 014-0)! x 51(7 — 5)!
4 7
=ixoay * Sn
myx LEEX SL 7X6 oy
BIx2x1. 2

Example 22 teachoo.com
Find the number of words with or without meaning which can be
made using all the letters of the word AGAIN. If these words are
written as in a dictionary, what will be the 50th word?
‘AGAIN’ = 2A, 1G, 11&1N
In dictionary, letters appear alphabetically,
Words starting | Representation Number of
with of words words
A Since, we arrange 4 letters,
A ‘—— Number of words = *P,
4 letters in
=4!=24
which A, G, |, N

Example 23 (Method 1) teachoo.com
How many numbers greater than 1000000 can be formed by using
the digits 1, 2,0, 2, 4, 2, 4?
There are total 7 digits in 1000000
We need to form a 7 digit number using the digits 1, 2, 0, 2, 4, 2, 4
But, these include numbers starting with ‘0’ like 0412224, ...etc
which are actually 6 digit numbers

Hence, we can’t have number beginning with 0
Thus,

Required numbers = All arrangements — Numbers starting with 0

Example 24 teachoo.com
In how many ways can 5 girls and 3 boys be seated in a row so that
no two boys are together?
The seating arrangement would be as follows
—~G_G_G_G_G_
5 girls can sit in any of 5 places 3 boys can sit at any of 6 places
marked —
Number of ways they can sit
sp Number of ways they can sit
=aTrs
=6
si = "Ps
~G-35)! _ 6}
5! 5! ~ (6-3)!
ort _6!_ 6X 5x 4x 3!
=5x4x3x2x1 ~ 317 31
= 120 =120

Misc 1 teachoo
How many words, with or without meaning, each of 2 vowels
and 3 consonants can be formed from the letters of the word
DAUGHTER?
DAUGHTER

Vowels Consonants

A,U,E D, G, H, T, R

Total number | Number to be Number of

chosen ways to choose
Vowels 3 2 3G,

Consonants 5 3 5C,

Misc 2 teachoo.com
How many words, with or without meaning, can be formed using all
the letters of the word EQUATION at a time so that the vowels and
consonants occur together?
EQUATION
Vowels Consonants
Number of vowels in EQUATION |Number of consonants in EQUATIO
=E,U,A,1,0 =Q,T,N
=5 =3
Number of ways vowels canbe |Number of ways consonants can be
arranged = °P,, arranged =°P,
_ sl 3!
~ (5-5)! ~ (3-3)!
515! 31 3!
== 7 7 120 =o 7778

Misc 3 teachoo.com
A committee of 7 has to be formed from 9 boys and 4 girls. In how
many ways can this be done when the committee consists of:
(i) exactly 3 girls?
Total | Number to | Number of ways
number | be chosen to choose
Girls 4 3 4c,
Boys 9 4 °C,
Total number of ways = “C, x °C,
4 91
= ———_—._ x ————_
34-3)! 49-4)!
4 91
=— x —
31! 41(5)!
91 9x8x7Xx6X5!
~ 3s)! (8x2x1)xG)! 504

Misc 4 teachoo.com
If the different permutations of all the letter of the word
EXAMINATION are listed as in a dictionary, how many words are
there in this list before the first word starting with E?
In dictionary, words are given alphabetically
We need to find words starting before E i.e. starting with A,B,C or D
In EXAMINATION, there is no B,C or D ,jhence words should start
with A
Words in the list before the word starting with E

= Words starting with letter A

Misc 5 (Method 1) teachoo.com
How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7
and 9 which are divisible by 10 and no digit is repeated?
A number is divisible by 10
120 is divisible by 10

if 0 is at the units place as last digit is 0
Thus, We need to form 6 digit number
whose unit place is 0

0
So,
We need to fill up 5 places with
the remaining digits 1, 3,5, 7,&9

Misc 6 teachoo.com
The English alphabet has 5 vowels and 21 consonants. How many
words with two different vowels and 2 different consonants can be
formed from the alphabet?
Total number | Number to be Number of
chosen ways to choose
Vowels 5 2 G
Consonants 21 2 21¢,
Number ways of selecting 2 vowels & 2consonants
=5C, x 21¢,
5! 21!
=——_ X ——_
25-2)! 221-2)!
5! 21!
= — x —
213! 21191

= SX4K3! 21x20 x19!

~2x1x3! 2x1x19!

=10x 210

= 2100

Misc 7 teachoo.com
In an examination, a question paper consists of 12 questions divided
into two parts i.e., Part land Part Il, containing 5 and 7 questions,
respectively. A student is required to attempt 8 questions in all,
selecting at least 3 from each part. In how many ways cana student
select the questions?
Student is required to attempt total 8 questions
& selecting atleast 3 question from part 1 & part 2
Atleast means minimum
So, she can select
Option 1 3 from Part 1 & 5 from Part 2) We have to
calculate all these
Option 2 4 from Part 1 & 4from Part 2 \ combinations
f f separately and
Option 3 5 from Part 1 & 3 from Part 2 then add it

Misc 8 teachoo.com
Determine the number of 5-card combinations out of a deck of 52
cards if each selection of 5 cards has exactly one king.
There are total 4 King Cards out of 52
We have to select 1 King from 4 King cards
The Remaining 4 we have to select from 48 cards (52 - 4 king cards)
Total | No. of card to | Number
be chosen of ways
King Cards 4 1 1,
Other Cards 48 4 8c,
Total number of ways = 4C, x °C,

Misc 9 teachoo.com
It is required to seat 5 men and 4 women in a row so that the women
occupy the even places. How many such arrangements are possible?
There are total 9 people
Since women occupy even places
Places 2, 4, 6, 8 will be occupied by women
And Rest will be occupied by men
The seating arrangement will be follows

EN CACCTIS

MW M WM WM WM

Misc 10 teachoo.com
From a class of 25 students, 10 are to be chosen for an excursion party.
There are 3 students who decide that either all of them will join or none
of them will join. In how many ways can the excursion party be chosen?
There are 2 options
All 3 join All 3 don’t join
Remaining 7 to be chosen Remaining 10 chosen from 22
from 25 — 3 = 22 students students
Number of ways = **C, Number of ways = ?7C,,
221 _ 22!
“712-7! ~ 101(22 — 10)!
- 22! 499 =—'_~ 646646
“Fas 170544 10112!

Misc 11 teachoo.com
In how many ways can the letters of the word ASSASSINATION be
arranged so that all the S’s are together?
Since we need to assigned 4S together,
We consider 4S as one block |SSSS
So, our letters become | SSSS ['] [N] ['] [o] [N]
We arrange them now
Since letters are repeating
Hence we use this formula = —

P,!P,1P5!