1.Algebra Booster

Permutations and Combinations 5.23
HINTS AND SOLUTIONS
LEVEL I
1. We have,
x x 1
+ =
5! 6! 7!
x x 1
fi + =
5! 6¥ 5! 7¥ 6¥
5!
x 1
fi x + =
6 7 ¥ 6
fi 7 x = 1
6 7 ¥ 6
1
fi x = .
49
2. We have,
(2 n)!
n!
2 n(2n-1)(2n-
2)…6.5.4.3.2.1
=
n!
{2.4.6…(2 n- 2)2 n} ¥ {1.3.5…(2 n-1)}
=
n!
n
2 {1.2.3…( n-1) ◊ n} ¥ {1.3.5…(2 n-1)}
=
n!
n
2 ¥ ( n)! ¥ {1.3.5…(2 n-1)}
=
n!
n
= 2 ¥ {1.3.5…(2 n -1)}
3. We have,
3.6.9.12…(3n
- 3)3n
n
3
n
3 {1.2.3…( n - 1) n}
=
n
3
n
3 ¥ ( n)!
=
n
3
= ( n)!
4. Since the unit digit of a factorial more than 4 is zero, so
the unit digit of the given expression
= 1! + 2! + 3! + 4! + … + (10)!
= Unit digit of 1! + 2! + 3! + 4!
= Unit digit of (1 + 2 + 6 + 24)
= Unit digit of 33
= 3
5. As we know that 5! = 120 and 4! = 24
Thus, e must be 4
Therefore, N = a! + b! + c! + d! + e!
= 0! + 1! + 2! + 3! + 4!
= 1 + 1 + 2 + 6 + 24
= 34
6. We have 1! ¥ 2! = 2
Thus, the maximum value of n is 2.
7. We have,
N = 1! + 2! + 3! + 4! + 5! + 6!
= 1 + 2 + 6 + 24 + 120 + 720
= 873
Thus, the maximum value of N is 6.
8. We have N = a! + b! + c!
Since the unit digit of N is 9, then
a = 1, b = 2 and c = 3
Therefore,
{a! ¥ b! ¥ c!}= {1! ¥ 2! ¥ 3!}
= {1 ¥ 1 ¥ 2 ¥ 1 ¥ 2 ¥ 3}
= 12
9. Since the last two digits of all the factorials more than
9 is 00, so the sum of all prime factorials
= 2! + 3! + 5! + 7!
= 2 + 6 + 120 + 5040
= 5168
Thus, the last two digits is 68.
10. Since x + y + z = 25, so the possible values of xyz = 9,
9, 7 and 9, 8, 8. Hence,
the required sum
= 3 ¥ 9! ¥ 9! ¥ 7! + 3 ¥ 9! ¥ 8! ¥ 8!
= 3 ¥ 9! ¥ 9 ¥ 8! ¥ 7! + 3 ¥ 9! ¥ 8! ¥ 8 ¥ 7!
= 3 ¥ 9! ¥ 8! ¥ 7!(9 + 8)
= 3 ¥ (9 + 8) ¥ 9! ¥ 8! ¥ 7!
= 51 ¥ 9! ¥ 8! ¥ 7!
11. We have,
1.1! + 2.2! + 3.3! + 4.4! … + n.n!
= (2 – 1) .1! + (3 – 1) .2! + (4 – 1) .3! + …
+ (n + 1) .n!
= (2! –1!) + (3! – 2!) + (4! – 3!) + …
+ ((n + 1)! – n!)
= ((n + 1)! – 1!)
= (n + 1)! – 1
12. We have,
1 2 3 4 99
+ + + + +
2! 3! 4! 5! 100!
(2 -1) (3 -1) (4 -1)
= + +
2! 3! 4!
(5 -1) (100 -1)
+ + +
5! 100!
Ê 1ˆ Ê 1 1ˆ Ê 1 1ˆ Ê 1 1 ˆ
= Á1
- + - + - + + -
Ë
˜
2! ¯
Á
Ë
˜
2! 3! ¯
Á
Ë
˜ Á ˜
3! 4! ¯ Ë99! 100! ¯
Ê 1 ˆ
= Á1 -
Ë
˜ 100! ¯