Example 1
A letter lock consists of three rings each marked with 10 different letters. In how many ways is it possible to make an unsuccessful attempt to open the block?
Solution:
Two rings may have same letter at a time but same ring cannot have two letters at time, therefore, we must proceed ring wise.
Each of the three rings can have any one of the 10 different letters in 10 ways.
Therefore Total number of attempts = 10 × 10 × 10 = 1000.
But out of these 1000 attempts only one attempt is successful.
Therefore Required number of unsuccessful attempts
= 1000 - 1 = 999.
Example 2
Find the total number of signals that can be made by five flags of different colour when any number of them may be used in any signal.
Solution:
Case I : When only one flag is used.
No. of signals made = 5P1 = 5.
Case II : When only two flag is used.
Number of signals made = 5P2 = 5.4 = 20.
Case III : When only three flags are used.
Number of signals is made = 5P3 = 5.4.3 = 60.
Case IV : When only four flags are used.
Number of signals made = 5P4 = 5.4.3.2 = 120.
Case V : When five flags are used.
Number of signals made = 5P5 = 5! = 120.
Hence, required number = 5 + 20 + 60 + 120 + 120 = 325.
Example 3
Prove that if each of the m points in one straight line be joined to each of the n points on the other straight line, the excluding the points on the given two lines. Number of points of intersection of these lines is 1/4 mn
(m-1(n-1).
Solution:
To get one point of intersection we need two points on the first line and two points on the second line. These can be selected out of n-points in nC2 ways and for m points in mC2 ways.
Therefore Required number = mC2 × nC2
= (m(m-1))/2! . (n(n-1))/2!
= 1/4 m n (m - 1)(n - 1)
Example 4
There are ten points in a plane. Of these ten points four points are in a straight line and with the exception of these four points, no other three points are in the same straight line. Find
(i) the number of straight lines formed.
(ii) the number of triangles formed.
(iii) the number of quadrilaterals formed by joining these ten points
Solution:
(i) For straight line, we need 2 points
No. of point selected out of 4 collinear points |
No. of points selected out of remaining 6 points |
No. of straight line formed |
0
1
2 |
2
1
0 |
4C0 × 6C2 = 15
4C1 × 6C1 = 24
1 |
(In last case only one straight line is formed)
Therefore Required number = 15 + 24 + 1 = 40
(ii) For triangle, we need 3 points
No. of point selected out of 4 collinear points |
No. of points selected out of remaining 6 points |
No. of triangles formed |
0
1
2
3 |
3
2
1
0 |
4C0 × 6C3 = 20
4C1 × 6C2 = 60
4C2 × 6C1 = 36
0 |
(In last case number of triangles formed is 0)
Therefore Required number = 20 + 60 + 36 + 0 = 116
(iii) For a quadrilateral, we need 4 points
No. of point selected out of 4 collinear points |
No. of points selected out of remaining 6 points |
No. of quadrilateral formed |
0
1
2
3
4 |
4
3
2
1
0 |
4C0 × 6C4 = 15
4C1 × 6C3 = 80
4C2 × 6C2 = 90
0
0 |
(In these cases number quadrilateral is formed
Therefore Required number = 15 + 80 + 90 = 185.
Example 5
A student is allowed to select at most n-blocks from a collection of (2n + 1) books. If the total number of ways in which he can select a book is 63, find the value of n.
Solution:
Since the student is allowed to select at the most n-books out of (2n + 1) books, therefore, he can choose, one book, two books or at the most n books. The number of ways of selecting at least one books are
2n+1C1 + 2n+1C2 + ......... 2n+1Cn = 63 = S (Say)
Again, we know that
2n+1C0 + 2n+1C1 + ......... 2n+1Cn + 2n+1C2n+1 = 22n+1
Now 2n+1C0 = 2n+1C2n+1 = 1
2n+1C1 = 2n+1C2n etc.........
Hence, we have
1 + 1 + 2S = 22n+1
or 2 + 2 . 63 = 22n+1
or 128 = 22n+1 or 27 = 22n+1
=> 2n+1 = 7
or 2n = 6
n = 3
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