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```					mathcity.org                                     Exercise 7.4 (Solutions)
Textbook of Algebra and Trigonometry for Class XI
Merging man and maths                     Available online @ http://www.mathcity.org, Version: 1.0.2

Question # 1 (i)
12!        12! 12 ⋅ 11 ⋅ 10 ⋅ 9! 12 ⋅ 11 ⋅ 10 1320
12
C3 =             =       =               =            =     = 220 Answer
(12 − 3)! 3! 9! 3!          9! 3!         3!        6
20!           20! 20 ⋅ 19 ⋅ 18 ⋅ 17! 20 ⋅19 ⋅ 18 6840
(ii)     20
C17 =                   =       =                 =            =     = 1140 Answer
( 20 − 17 )!17! 3!17!            3!17!            3!       6
n!        n (n − 1)(n − 2)(n − 3)(n − 4)! n (n − 1)(n − 2)(n − 3)
(iii)    n
( n − 4 )! 4!             ( n − 4 )! 4!                     4!

Question # 2 (i)
Since nC5 = nC4
⇒ nCn −5 = nC4                          Q nCr = nCn−r
⇒ n−5=4                ⇒ n= 4+5                ⇒ n=9
12 × 11                     12 ⋅ 11 ⋅ 10!                        12!
(ii)            n
C10 =                    ⇒ nC10 =                       ⇒ nC10 =
2!                          2!10!                         (12 − 10)!10!
⇒ nC10 = 12C10                 ⇒ n = 12 .
(iii)
Do yourself as Q # 2 (i)

Question # 3 (i)
Cr = 35 and n Pr = 210
n

n!                   n!
Since nCr = 35         ⇒                   = 35 ⇒             = 35 ⋅ r ! ………. (i)
( n − r )! r !         ( n − r )!
n!
Also n Pr = 210 ⇒                      = 210 ………… (ii)
( n − r )!
Comparing (i) and (ii)
35 ⋅ r ! = 210
210
⇒ r!=                ⇒ r ! = 6 ⇒ r ! = 3! ⇒ r = 3
35
Putting value of r in equation (ii)
n!
= 210
( n − 3)!
n (n − 1)(n − 2)(n − 3)!
⇒                               = 210
( n − 3 )!
⇒ n (n − 1)(n − 2) = 210 ⇒ n (n − 1)(n − 2) = 7 ⋅ 6 ⋅ 5
⇒     n= 7

(ii)                       n−1
Cr −1 : nCr : n+1Cr +1 = 3: 6 :11
n−1                                     (n − 1)!               n!
First consider           Cr −1 : nCr = 3: 6      ⇒                              :               = 3:6
( n − 1 − r + 1)! (r − 1)! ( n − r )! r !
(n − 1)!

⇒
(n − 1)!
:
n!
= 3: 6        ⇒
( n − r )! (r − 1)! = 3
( n − r )! (r − 1)! ( n − r )! r !                          n!           6
( n − r )! r !
FSc-I / Ex 7.4 - 2

⇒
(n − 1)!
×
( n − r )! r ! = 1      ⇒
(n − 1)! r ! 1
×      =
( n − r )! (r − 1)!         n!          2            (r − 1)! n! 2
(n − 1)! r (r − 1)! 1                  r 1
⇒              ×              =         ⇒       =         ⇒ n = 2r ………. (i)
(r − 1)! n (n − 1)! 2                  n 2
Now consider
n!                 (n + 1)!
n
Cr : n +1Cr +1 = 6 :11 ⇒                     :                           = 6 :11
( n − r )! r ! ( n + 1 − r − 1)! (r + 1)!
n!

⇒
n!
:
(n + 1)!
= 6 :11   ⇒
( n − r )! r !     =
6
( n − r )! r ! ( n − r )! (r + 1)!                    (n + 1)!             11
( n − r )! (r + 1)!
⇒
n!
×
( n − r )! (r + 1)! = 6       ⇒
n ! (r + 1)! 6
×        =
( n − r )! r !        (n + 1)!         11         r ! (n + 1)! 11
n ! (r + 1) r ! 6                     (r + 1) 6
⇒    ×                =           ⇒            =       ⇒ 11(r + 1) = 6(n + 1)
r ! (n + 1) n! 11                    (n + 1) 11
⇒ 11(r + 1) = 6(2r + 1)                       Q n = 2r
⇒ 11r + 11 = 12r + 6 ⇒ 11r − 12r = 6 − 11
⇒ − r = −5 ⇒           r =5
Putting value of r in equation (ii)
n = 2(5)           ⇒ n = 10

Question # 4 (i)
(a) 5 sided polygon has 5 vertices,
so joining two vertices we have line segments = 5C2 = 10
Number of sides = 5
So number of diagonals = 10 – 5 = 5
(b) 5 sided polygon has 5 vertices,
so joining any three vertices we have triangles = 5C3 = 10
Question # 4 (ii)
(a) 8 sided polygon has 8 vertices
So joining any two vertices we have line segments = 8C2 = 28
Number of sides = 8
So number of diagonals = 28 – 8 = 20
(b) 8 sided polygon has 8 vertices,
so joining any three vertices we have triangles = 8C3 = 56 .
Question # 4 (iii)
Do yourself as above.

Queston # 5
Number of boys = 12
So committees formed taking 3 boys = 12C3 = 220
Number of girls = 8
So committees formed by taking 2 girls = = 8C2 = 28
Now total committees formed including 3 boys and 2 girls = 220 × 28 = 6160

Made by: Atiq ur Rehman ( atiq@mathcity.org ), URL: http://www.mathcity.org
FSc-I / Ex 7.4 - 3

Question # 6
Number of persons = 8
Since two particular persons are included in every committee so we have to find
combinations of 6 persons 3 at a time = 6C3 = 20
Hence number of committees = 20
Question # 7
The number of player = 15
So combination, taking 11 player at a time = 15C11 = 1365
Now if one particular player is in each collection
then number of combination = 14C10 = 1001

Question # 8
L.H.S = 16C11 + 16C10
16!           16!         16!    16!
=               +               =      +
(16 − 11)! 11! (16 − 10 )! 10! 5! 11! 6! 10!
16!          16!       16!  1 1 
=              +          =        + 
5! 11 ⋅ 10! 6 ⋅ 5! 10! 10! 5!  11 6 
16!  6 + 11       16!  17     16!  17 
=                =         =                
10! 5!  66  10! 5!  66  10! 5!  11 ⋅ 6 
17 ⋅ 16!       17!          17!
=                =        =              = 17C11 = R.H.S
11 ⋅ 10! 6 ⋅ 5! 11! 6! 11! (17 − 11)!
Alternative
L.H.S = 16C11 + 16C10 = 4368 + 8008 = 12276 …..….. (i)
R.H.S = 17C11 = 12376 ……… (ii)
From (i) and (ii)
L.H.S = R.H.S

Question # 9
Number of men = 8
Number of women = 10
(i) We have to form combination of 4 women out of 10 and 3 men out of 8
= 10C4 × 8C3 = 210 × 36 = 11760
(ii) At the most 4 women means that women are less than or equal to 4, which implies
the following possibilities (1W ,6M ) , ( 2W ,5M ) , ( 3W ,4M ) , ( 4W ,3M ) , ( 7 M )
= 10C1 × 8C6 + 10C2 × 8C5 + 10C3 × 8C4 + 10C4 × 8C3 + 8C7
= (10 )( 28) + ( 45 )( 56 ) + (120 )( 70 ) + ( 210 )( 56 ) + ( 8)
= 280 + 2520 + 8400 + 11760 + 8
= 22968
(iii) At least 4 women means that women are greater than or equal to 4, which implies
the following possibilities ( 4W ,3M ) , ( 5W ,2M ) , ( 6W ,1M ) , ( 7W )
= 10C4 × 8C3 + 10C5 × 8C2 + 10C6 × 8C1 + 10C7
= ( 210 )( 56 ) + ( 252 )( 28) + ( 210 )(8 ) + 120
= 11760 + 7056 + 1680 + 120
= 20616
FSc-I / Ex 7.4 - 4

Question # 10
n!                   n!
L.H.S = nCr + nCr −1 =                 +
( n − r )! r ! ( n − (r − 1) )! ( r − 1)!
n!                  n!
=                +
( n − r )! r ! ( n − r + 1)! ( r − 1)!
n!                           n!
=                       +
( n − r )! r ( r − 1)! ( n − r + 1)( n − r )! ( r − 1)!
n!        1          1      
=                         +              
 r ( n − r + 1) 
( n − r )! ( r − 1)!
n!           n − r +1+ r 
=                                      
( n − r )! ( r − 1)!  r ( n − r + 1) 
n!             n +1 
=                                     
 r ( n − r + 1) 
( n − r )! ( r − 1)!
=
( n + 1) n!
( n − r + 1)( n − r )! r ( r − 1)!
=
( n + 1)!        =
( n + 1)!
( n − r + 1)!   r!       ( n + 1 − r )!   r!
= n +1Cr = R.H.S

Made by: Atiq ur Rehman ( atiq@mathcity.org ), URL: http://www.mathcity.org

Submit error/mistake online at http://www.mathcity.org/error

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