1.Algebra Booster

Permutations and Combinations 5.15
3. Find the number of positive integral values of x for
which 4( x – 1 C 4
– x – 1 C 3
) < 5(x – 2)(x – 3).
4. If m is the number of ways, where 4 boys and 4 girls
are seated in a row so that they are alternate and n is the
number of ways, where 5 boys and 4 girls are seated in
a round table so that they are alternate, find the value of
Êm
ˆ
Á + 2
Ë
˜
n ¯ .
5. Let P n
denotes the number of ways in which three people
can be selected out of n people sitting in a row so
that no two of them are consecutive.
If P n + 1
– P n
= 15, find n.
n
n
P
n
r- 1 P P
r r+
6. If = = 1 , find the value of
a b c
Ê
2
b ˆ
Á + 2
Ëab
( + c)
˜
¯ .
7. If m be the number of ways, in which 7 persons can be
seated in a round table if 2 particular persons may not
sit together and n the number of arrangement of letters
of the word DELHI, where E always comes before I,
Êm
ˆ
find the value of Á - 3
Ë
˜
n ¯ .
8. If the number of ordered triplets of positive integers
which satisfy the inequality 11 £ (a + b + c) £ 50 is
Ê x ˆ
( x C 3
– y C 3
), find the value of Á + 2
Ë y
˜
¯ .
9. If m be the number of positive integral solutions of
xyz = 30 and n the number of integral solutions of abcd
= 210 such that m = 3 p and n = 2 q , find the value of
(q – 2p).
10. In how many ways, 5 identical balls can be distributed
into 3 different boxes so that no box remain empty?
11. If m be the number of ways in which a score of 11 can
be made from a throw by three persons, each throwing
a single die once and n is the number of ways 5 apples
be distributed among the 3 students, so that each can
get any number of apples, find the value of (m – n).
12. If m be the number of different words that can be made
from the word BHARAT in which B and H are never
together and n be the number of words that can be
made from the letter of the word LAUGH if vowels
Êm
ˆ
occur together, find the value of Á + 3
Ë
˜
n ¯ .
13. If the number of ways in which n distinct objects can
be put into two identical boxes, so that no box remains
empty, is 127, find n.
n
14. Find the number of values of n, for which  ( k!)
is
the square of an integer.
k = 1
15. In a certain test, there are n questions. In this test, 2 n–k
students gave wrong answers for at least k questions,
where k = 1, 2, 3, …, n. If the total number of wrong
answers given is 511, find the value of n.
16. Everybody in a room shakes hands with everybody
else. If the total number of handshakes is 36, find the
number of people in the room.
17. In a bakery shop, four types of biscuits are available.
If a person can buy 10 biscuits, if he decides to take
at least one biscuit of each variety is x C y
ways, find the
value of (x – y – 2).
18. In a JEE-Advanced mock test, there are n questions. In
this test, 3 n–k students gave wrong answers for at least k
questions, where k = 1, 2, 3, …, n. If the total number
of wrong answers is 3280, find the value of n.
19. There are 4 pairs of hand gloves of 4 different colours.
In how many ways can they be paired off so that a left
handed glove and a right handed glove are not of the
same colour?
20. If the number of possible outcomes in a throw of n ordinary
dice in which at least one of the dice shows an
odd number is 189, find the value of n.
Comprehensive Link Passages
Passage I
Let p be a prime number and n a positive integer, the exponent
of a prime p in n! is E p
(n!) and is given by
Èn˘ È n ˘ È n ˘ È n ˘
Ep
( n!)
= Í 2 3
k
p ˙+ Í + + +
p
˙ Í
p
˙ Í
p
˙
Î ˚ Î ˚ Î ˚ Î ˚
where p k < n < p k + 1 , and [] = GIF
If we isolate the power of each prime contained in any
a
number N, N can be written as
1 a
2 3
2 a
N 5
3 a
= ◊ ◊ ◊ 7
4,
where
a i
are whole numbers.
On the basis of the above information, answer the following
questions.
1. The exponent of 7 in 100 C 50
is
(a) 0 (b) 1 (c) 2 (d) 3
2. The number of zeroes at the end of 108! is
(a) 10 (b) 13 (c) 25 (d) 26
3. The last non-zero digit in 20! must be equal to
(a) 2 (b) 4 (c) 6 (d) 8
4. The exponent of 12 in 100! is
(a) 32 (b) 48 (c) 97 (d) none
5. The number of prime numbers among the numbers
(105)! + 2, (105)! + 3, (105)! + 4, …, (105)! + 104 is
(a) 30 (b) 32 (c) 33 (d) none
Passage II
Suppose a lot of n objects contains n 1
objects of one kind, n 2
objects of second kind, n 3
objects of third kind, …, n k
objects
of kth kind, such that n 1
+ n 2
+ … + n k
= r.
The number of possible arrangements of r objects
out of this lot is the co-efficient of x r in the expansion of
Ê
n1
Ê
l
x ˆˆ
()! r ¥ ’ ÁÂ
Á ˜
l = 0Ë( l)!
˜
.
Ë ¯¯