1.Algebra Booster

Permutations and Combinations 5.45
No. of signals using 4 flag = 5 P 4
No. of signals using 5 flag = 5 P 5
Hence, the total number of ways
= 5 P 1
+ 5 P 2
+ 5 P 3
+ 5 P 4
+ 5 P 5
= 5 + 20 + 60 + 120 + 120
= 325
2. The rank of the word TOUGH in dictionary,
m = 4! + 4! + 4! + 3! + 3! + 2! + 2! + 1
= 89
Clearly, the rank of the word IIT is 1
Thus, n = 1
Hence, the value of m + n + 10
= 89 + 1 + 10
= 100
3. The required number of triangles
= n C3
- n - n( n - 4)
nn ( -1)( n-
2)
= - n - n( n - 4)
6
n 2
= ( n - 3 n + 2- 6- 6 n + 24)
6
n 2
= ( n - 9 n + 20)
6
nn ( - 4)( n-5)
=
6
4. Let S 1
, S 2
, S 3
, …, S 8
denote the stations where the train
does not stop.
So, there are four stations where the train stops.
It can be possible in 9 C 4
ways
= 9 ¥ 8 ¥ 7 ¥ 6 = 126 ways.
24
5. A pack of cards consists of 4 suits, namely, spade, club,
diamond and hearts, respectively. Each suit consists of
13 cards. Number of ways of choosing 5 cards in 9 different
ways.
But one card of any denominations can be selected
from 4 suits in 4 5 ways.
Hence, the number of ways a hand containing 5 consecutive
denominations
= 9 ¥ 4 5 = 9 ¥ 1024 = 9216
6. To get one point of intersection, we take 2 points on the
first line and 2 points on the second line.
It can be done in m C 2
¥ n C 2
ways
mm ( -1) nn ( -1)
= ¥
2 2
mn( m -1)( n -1) =
ways.
4
7. Let the number of children of John and his first wife be
x and the number of children of John and Mary be y.
So, the number of children of Mary and her first husband
= x + 1.
Thus, x + x + 1 + y = 24
2x + y = 23
Total number of fights between two children
= 24 C 2
= 276.
Now, the total number of fights between two children
of same parents = x + 1 C 2
+ x C 2
+ y C 2
= ( x C 1
+ x C 3
) + x C 2
+ y C 2
= ( x C 1
+ 2 x C 2
) + 23 – 2x C 2
= 3x 2 – 45x + 253
Therefore, the total number of fights, subject to the
condition that any two children of the same parents do
not fight.
N = 276 – (3x 2 – 45x + 253)
fi N = 23 – 3x 2 + 45x
dN
fi 6x
45
dx =- +
dN
For maximum or minimum, 0
dx =
fi –6x + 45 = 0
fi x = 7 ◊ 5
fi x = 7
2
d N
Also, =- 6<
0
2
dx
Thus, N will be maximum, when x = 7.
Hence, the maximum number of fights
= 23 – 3 ¥ 7 2 + 45 ¥ 7
= 23 – 147 + 315
= 191
8. The possible triplets of 5 are (1, 2, 2) and (1, 1, 3).
Hence, the number of possible ways
5! 3! 5! 3!
= ¥ + ¥
1! ¥ 2! ¥ 2! 2! 1! ¥ 1! ¥ 3! 2!
= 30 ¥ 3 + 20 ¥ 3
= 90 + 60
= 150
9. The number of possible ways
= Co-efficient of x 16 in (x 3 + x 4 + ... + x 7 ) 4
= Co-efficient of x 16 in x 12 (1 + x + ... + x 4 ) 4
= Co-efficient of x 4 in (1 + x + ... + x 4 ) 4
= Co-efficient of
5
4
Ê1
- x ˆ
x inÁ
Ë 1 - x
˜
¯
= Co-efficient of x 4 in (1 – x 5 ) 4 ¥ (1 – x) –4
= Co-efficient of x 4 in (1 – x) –4
= Co-efficient of x 4 in
(1 + 4 C 1
x + 5 C 2
x 2 + 6 C 3
x 3 + 7 C 4
x 4 + ... )
7 7¥ 6¥ 5¥
4
= C4
= = 35
24
10. The number of possible ways
= Co-efficient of x 30 in (x 2 + x 3 + ... + x 16 ) 8
= Co-efficient of x 30 in x 16 (1 + x + ... + x 14 ) 8
= Co-efficient of x 14 in (1 + x + ... + x 14 ) 8
4