1.Algebra Booster

Permutations and Combinations 5.47
(1 – x) –1 (1 – x 2 ) –1 (1 – x 3 ) –1
= (1 + x + x 2 + x 3 + x 4 )(1 + x 2 + x 4 )(1 + x 3 )
= (1 + x + x 2 + x 3 + x 4 )(1 + x 2 + x 3 + x 4 )
= 1 + 1 + 1 + 1
= 4
19. Consider I for India and P for Pakistan.
We can arrange I and P to show wins for India and Pakistan
respectively.
Thus, IPPPP means first match won by India and 4
matches won by Pakistan, respectively.
Consider Pakistan wins the series, the last match is always
won by Pakistan
Wins of I Wins of P No. of ways
0 4 1
4!
1 4
3! = 4
2 4
3 4
5!
2!3! = 10
6!
3!3! = 20
Thus, the total number of ways = 35
Similarly, the same number of ways India can win the
series = 35
Therefore, total number of ways, the series can be won
= 35 + 35 = 70.
20. Find the number of quintuples (x, y, z, u, v) of positive
integers satisfying both.
x + y + z + u = 30 and x + y + z + v = 27
22. Given 3x + y + z = 24, x ≥ 0, y ≥ 0, z ≥ 0
Let z = k fi y + z = 24 – 3k
So, 0 £ 24 – 3k £ 24
fi –24 £ –3k £ 0
fi 0 £ k £ 8
Thus, the number of integral solutions
= 24–3k+2–1 C 2–1
= 25–3k C 1
= (25 – 3k)
Hence, the number of integral solutions
8
= (25 -3 k)
Â
k = 0
8 8
Â
= 25 -3
Â
k= 0 k=
0
k
3.8.9
= 25 ¥ 9 -
2
= 225 -108
= 117
23. The possible triplets to get 12 are (1, 5, 6), (2, 4, 6),
(3, 4, 5), (3, 3, 6), (2, 5, 5) and (4, 4, 4)
Hence, the number of possible ways
3! 3! 3!
= 3! + 3! + 3! + + +
2! 2! 3!
= 18 + 6 + 1
= 25.
24. The number of possible ways it can be done
= 2! ¥ (18)!
= 2 ¥ (18)!
25. Let the student get x i
marks in each paper.
Thus, x 1
+ x 2
+ x 3
+ x 4
= 2m, 0 £ x i
£ m
Therefore, the number of ways of getting 2m marks =
the number of non-negative integral solutions of the
equation.
= Co-efficients of x 2m in (1 + x + x 2 + x 3 + … + x m ) 4
= Co-efficients of x 2m in (1 – x m+1 ) 4 ¥ (1 – x) –4
= Co-efficients of x 2m in
(1 – x m+1 ) 4 ¥ (1 + 4 C 1
x + 5 C 2
x 2 + 6 C 3
x 3 + …)
= 2m+3 C 2m
– 4( m+2 C m–1
)
= 2m+3 C 3
– 4( m+2 C 3
)
26. We have
(2m+ 3)(2m+ 2)(2m+ 1) 4( m+ 2)( m+
1) m
= -
6 6
( m + 1)
= ((2 m+ 3)(2 m+ 1) - 2 m ( m+
2))
3
( m + 1) (4
2 8 3 2
2 4 )
= m + m+ - m - m
3
1 ( 1)(2
2
= m+ m + 4 m+
3)
3
2016
x= Â ( k!)
k = 1
= 1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9! + 10!
+ 11! + 12! + … + (2016!)
In 5! and after last digit is 0 and after (10!) last two
digit is 00
So sum of the last two digits
= 01 + 02 + 06 + 24 + 20 + 20 + 40 + 20 + 80
= 13
Hence, the last two digits in
2016
x= Â ( k!)
is 13.
k = 1
27. We select 4 digits out of 10 in 10 C 4
ways.
Since the order is fixed for arrangement, i.e.
a > b > c > d.
Hence, the number of possible ways
= (Number of ways to select) ¥ 1
= 10 C 4
¥ 1
10 ¥ 9 ¥ 8 ¥ 7
= = 210
24
28. (i) Consider two Bs as single object, the letters BB,
C, D in 3! ways.
i.e. ¥ BB ¥ C ¥ D ¥
So there are four gaps. We can fill As in four gaps
which is possible in 4 C 3
ways.
Hence, the required number of arrangements
= 4 C 3 ¥ 3!
= 4 ¥ 6 = 24