1.Algebra Booster

Permutations and Combinations 5.51
8. Two women can be seated in 4 P 2
ways and 3 men can
occupy the chairs in 6 P 3
in ways.
Thus, the total number of possible ways
= 4 P 2
¥ 6 P 3
9. As the number of students answering incorrect at least
r questions is a r
.
The number of students answering exactly r questions
[1 £ r £ (n – 1)] questions incorrect is a r
– a r–1
Also, the number of students answering all questions
wrong is a k
.
Thus, the total number of possible ways, a student can
give wrong answers is
= 1(a 1
– a 2
) + 2(a 2
– a 3
) + … + (k – 1)(a k–1
– a k
) + ka k
= a 1
+ a 2
+ … + a k
10. m men can take their seat in m! ways.
After m men have taken their seats, the women can take
their n seats out of (m + 1) seats.
It can be done in n C m + 1
ways.
\ Total possible ways
n
= m! ¥ Cm+
1 ¥ n!
m! ¥ ( m+
1)!
=
( m- n+
1)!
11.
12. Husband and wife can invite their relatives in (3M, 0F),
(2M, 1F), (1M, 2F) and (0M, 3F) ways
The total number of possible ways
= ( 4 C 3
¥ 4 C 3
) + ( 4 C 2
¥ 4 C 1
)( 3 C 1
¥ 4 C 2
)
+ ( 4 C 1
¥ 3 C 2
)( 3 C 2
¥ 4 C 1
) + ( 3 C 3
¥ 3 C 3
)
= 485
13. The number of ways of choosing 3 balls out of 2 white,
9
3 black and 4 red balls = C3
9¥ 8¥
7
= = 84
6
If no black ball is included, the number of possible
6
ways = C3
6¥ 5¥
4
= = 20
6
Thus, the number of ways at least one ball is included
in the selection = 84 – 20
= 64
14. We have
2n+1
C 1
+ 2n+1 C 2
+ 2n+1 C 3
+ … + 2n+1 C n
= 3
fi
2n+1 C 0
+ 2n+1 C 1
+ 2n+1 C 2
+ 2n+1 C 3
+ … + 2n+1 C n
= 63 + 2n+1 C 0
= 64
1 2n+
1
fi ¥ 2 = 64
2
fi 2 2n = 64 = 2 6
fi 2n = 6
fi n = 3
15. We can arrange 6 ‘+’ signs in just one way.
Then we will have 7 gaps.
In these 7 gaps, 4 ‘–’ can be placed in 7 P 4
ways.
Thus, the total possible ways
7
= 6! ¥ C
4!
4 ¥
6! 4!
= 35
16 As 0 + 1 + 2 + 3 + 4 + 5 = 15 to form a 5-digit number
divisible by 3, we must leave either 0 or 3.
When 0 is left out, the numbers are 5 P 5
When 3 is left out, then numbers are 5 P 5
– 4 P 4
Also, 0 cannot be used at extreme left.
Thus, the required number of ways
= 5 P 5
+ 5 P 5
– 4 P 4
= 120 + 120 – 24
= 240 – 24
= 216
17.
18. Note that 4 + 3 = 7 guests have already selected the
sides where they wish to sit. We can choose 5 guests for
the particular side in 11 C 5
ways.
Now the 9 guests can be arranged on the particular
side in 9 P 9
ways and also 9 guests on either sides in 9 P 9
ways.
Thus, the total number of possible ways
= 11 C 5
¥ 9! ¥ 9!
19. Hence, the number of possible ways
Ê 1 1 1 1ˆ
= 4! ¥ Á1 - + - +
Ë 1! 2! 3! 4! ˜
¯
Ê 1 1 1ˆ
= 4! ¥ Á - +
Ë2! 3! 4! ˜
¯
Ê12 - 4 + 1ˆ
= 4! ¥Á
Ë
˜
24 ¯
Ê 9 ˆ
= 24 ¥Á
Ë
˜
24¯
= 9
20. Method Women Men Number of ways
I 5 7 9
C 5
¥ 8 C 7
= 1008
II 6 6 9
C 6
¥ 8 C 6
= 2352
III 7 5 9
C 7
¥ 8 C 5
= 2016
IV 8 4 9
C 8
¥ 8 C 4
= 630
V 9 3 9
C 9
¥ 8 C 3
= 56
Thus, the women in majority in III, IV and V.
So, the possible ways the women in majority
= 2016 + 630 + 56
= 2702
Also, the men in majority in I
Thus, the total possible ways = 1008.
21.
22.