1.Algebra Booster

5.48 Algebra Booster
(ii) The letters 2Bs, 1 C, 1 D can be arranged in
4! = 24 = 12 2! 2
Consider the arrangement is
¥ B ¥ B ¥ C ¥ D ¥
Clearly, there are 5 gaps. In these gaps, we will fill
3As in 5 C 3
ways.
Hence, the required number of arrangements
= ( 5 C 3
¥ 12 – 24)
= (10 ¥ 12 – 24)
= 120 – 24
= 96
29. Given,
P n
– Q n
= 6
fi n–2 C 3
– ( n C 3
– n(n – 3)) = 6
fi n–2 C 3
– n C 3
+ n(n – 3) = 6
fi ( n–2 C 3
– n C 3
) + n(n – 3) = 6
Ê( 2)( 3)( 4) ( 1)( 2)
fi
n - n - n - n n - n - ˆ
Á
-
Ë
˜
6 6 ¯
= n(n – 3) = 6
fi (n –2)[(n – 3)(n – 4) – n(n – 1)] + 6n(n – 3) = 36
fi –(n –2)(6n – 12) + 6n(n – 3) = 36
fi –(n –2)(n – 2) + n(n – 3) = 6
fi –n 2 + 4n – 4 + n 2 – 3n = 6
fi n – 4 = 6
fi n = 10
Hence, the value of n is 10.
30. A function f: A Æ B such that
f(0) £ f(1) £ f(2)
is defined as follows
Case I: When f(0) < f(1) < f(2)
Number of functions = 8 C 3
Case II: When f(0) = f(1) < f(2)
Number of functions = 8 C 2
Case III: When f(0) < f(1) = f(2)
Number of functions = 8 C 2
Case IV: When f(0) = f(1) = f(2)
Number of functions = 8 C 1
Hence, the total number of functions
= 8 C 3
+ 8 C 2
+ 8 C 2
+ 8 C 1
= 56 + 28 + 28 + 8 = 120
Integer Type Questions
1. We have
n+1
C r+1
: n C r
: n–1 C r–1
= 11:6:3
fi
fi
n+
1
C
C
n
r + 1
r
n
11 Cr
6
= , =
n-1
6 C 3
Ên+
1ˆ 11 n
Á = , = 2
Ër
+ 1˜
¯ 6 r
r-1
fi 11r + 11 = 6n + 6, n = 2r
fi 11r + 11 = 12r + 6
fi r = 5
2. We have
n
n
C 7 C
r
r + 1 3
= and =
n
n
C 3 C 2
r-1
r
1 7 3
fi
n - r +
and
n -
= r =
r 3 r + 1 2
fi 10r – 3n = 3 and 2n – 2r = 3r + 3
fi 10r – 3n = 3 and 2n = 5r + 3
fi 10r – 3n = 3 and 4n = 10r + 6
fi 4n – 6 – 3n = 3
fi n = 9
Hence, the value of n is 9.
3. We have
4( x–1 C 4
– x–1 C 3
) < 5(x – 2)(x – 3)
( 4) 1
fi
Ê x - ˆ
4( x-1)( x-2)( x-3)
Á -
Ë
˜
24 6¯
< 5( x-2)( x-3)
( x - 8)
fi 4( x-1)( x-2)( x- 3) < 5( x-2)( x-3)
24
fi (x – 1)(x – 3)(x – 8) < 30(x – 2)(x – 3)
fi (x – 2)(x – 3){(x – 1)(x – 8) – 30} < 0
fi (x – 2)(x – 3){(x 2 – 9x – 22) < 0
fi (x – 2)(x – 3)(x – 11)(x + 2) < 0
fi –2 < x < 2, 3 < x < 11
\ x = 1, 4, 5, 6, 7, 8, 9, 10
x = 5, 6, 7, 8, 9, 10, since x = 1 and 4 do not satisfy the
given in equation.
4. Clearly, m = 2! ¥ 4! ¥ 4!
and n = 4! ¥ 4!
Êm
ˆ
Hence, the value of Á + 2
Ë
˜
n ¯ = 4.
5. We have,
n–1
C 3
– n–2 C 3
= 15
fi ( n- 1)( n- 2)( n- 3) ( n- 2)( n- 3)( n-
- 4) = 15
6 6
fi
( n-2)( n-3) ( n- 1 - n+ 4) = 15
6
fi (n – 2)(n – 3) = 30
fi n 2 – 5n + 6 = 30
fi n 2 – 5n – 24 = 0
fi (n – 8)(n + 3) = 0
fi n = 8, –3
fi n = 8
Hence, the value of n is 8.
6. We have
n
n
P
n
r- 1 P P
r r+
1
= =
a b c
fi
n
n
b P P
r c r + 1
= , =
n
n
a P b P
r-1
r