1. There are n lines drawn in a plane such
that no two of them are parallel and no
three of them are concurrent. The number
of different points at which these lines will
cut is ?
2. Number of 6-digit telephone numbers, which
can be constructed with digits 0, 1, 2, 3,
4, 5, 6, 7, 8, 9, if each number starts with
35 and no digit appears more than once is
(A) 1680 (B) 8! (C) 6! (D) 6.6!
3. In a plane there are 37 straight lines of
which 13 pass through the point A and 11
pass through the point B. Besides, no three
lines pass through one point, no line passes
through both points A and B, and no two
are parallel. Then the number of intersec-
tion points the lines have is equal to
(A) 535 (B) 601
(C) 728 (D) None of these
4. From six gentlemen and four ladies, a
commitete of five is to be formed. Number
of ways in which this can be done if the
committee is to include at least one lady
is ?
5. Four students of class X, five students of
class XI and six students of class XII sit in
row. The number of ways, they can sit in
a row so that students belonging to the
same class are together is?
6. The number of rectangles exculding squares
from a rectangle of size 9 × 6 is?
7. A teaparty is arranged of 16 people along
two sides of a large table with 8 chairs on
each side. Four men want to sit on one
particular side and two on the other side.
The number of ways in which they can be
seated is
8. A five digit number divisible by 3 is to be
formed using the numerals 0, 1 , 2, 3, 4
and 5 without repetition. The total number
of ways this can be done is
9. Given that n is odd, the number of ways
in which three numbers in A.P. can be
selected from, 1, 2, 3,....n is ?
10. The number of ways in which 52 cards can
be divided into 4 sets, three of them hav-
ing 17 cards each and the fourth one having
just one card ?
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12 Answers
9 th sum i san awesome sum ....
hint the trick lies in selecting the middle term..... got it (question copied from titiu aderseecus 102 combinatorics.....)
10) there may be a problem in ordering of sets...... means here all the 3 sets are identical........
5)
Treating them as three groups, the three groups can be arranged in 3
ways. Thereafter each group can be arranged inter-nally. Total number
= 3! × 4! × 5! × 6!
8)
If a number is divisible by 3, the sum of the digits in it must be multiple
of 3. The sum of the given six numerals is 0 + 1 + 2 + 3 + 4 + 5 = 15.
So to make a five digit number divisible by 3 we can either exclude 0 or
3. If 0 is left out, then 5! = 120 number of ways are possible. If 3 is left
out, then the number of ways of making a five digit numbers is 4 × 4!
= 96, because 0 cannot be placed in the first place from left, as it will
give a number of four digits. Hence, total number = 120 + 96 = 216.
2)
Since each number consisting of 6 digits starts with 35, 3 and 5 are
fixed in the first and second places. The other four places can be filled
up with remaining 8 digits in 8P4 ways = 8.7.6.5 =
1680 ways. Hence the required numbers of telephone number is 1680.
4)
No. of committees = (6C4 x 4C1) + (6C3 x 4C2) + (6C2 x 4C3) + (6C1 x 4C4)