Example.1
How many 4-letter
words can be formed using a, b, c, d and e
(i) without
repetition
(ii) with
repetition.
Solution
(i) The number of words that can be formed
is equal to the number of ways of filling the four places.
&⇒ 5´4´3´2=120
words can be formed when repetition is not allowed.
(ii) The number of words that can be formed
is equal to the number of ways of filling the four places.
First place can be
filled in 5 ways. If repetition is allowed, all the remaining places can be
filled in 5 ways each.
&⇒ 5´5
´ 5 ´ 5 = 625 words can be formed when
repetition is allowed.
Example.2
(a) In how many ways can we form 8-digit
numbers from the digits
0, 1, 2, 3, 4, 5, 6, 7? (Repetition not allowed)
(b) How many of these are even ?
(c) How many of these are divisible by 25?
(d) How many of these are less than 20000
?
(e) Find the number of integers greater
than 7000 that can be formed with the digits 3, 5, 7, 8 and 9, no digit is
being repeated. .
Solution
(a) There are 8
places to be filled.
First place can not
be filled by 0. I.e. First place can be filled in 7 ways. 2nd place
can be filled either by the 6 remaining digits or by 0 i.e. in 7 ways. 3rd
place can be filled in 6 ways (by the 6 remaining digits) and so on.
7 ´ 7 ´ 6 ´
5 ´ 4 ´ 3 ´ 2 ´
1 = 7 ´ 7!
Alternative :
Total no. of
possible permutations = 8! Of these, there are some permutations in which 0
comes in the first place i.e. they are actually 7 digit numbers. No. of such
permutations = 1 ´ 7! (there is only 1
choice for the first place i.e. 0).
So that no. of
8-digit numbers = 8! - 7! = 7! (8 - 1) = 7 ´ 7!.
(b) Even nos. can be categorised into two
categories.
(i) Last digit is 0.
For the remaining 7 digits, there are 7 choices .
&⇒ 7! ways.
(ii) Last digit is
not 0. For the last place, there are 3 choices i.e. 2, 4 and 6. There are 6
choices for the 1st place (excluding zero and the digit in last
place).
There are 6, 5, 4,
3, 2, and 1 choices for the middle positions (i.e. remaining digits).
So no. of such
numbers = 6 ´ 6 ´ 5 ´ 4 ´
3 ´ 2 ´ 1 ´ 3 = 18 ´
6!.
So total no. of
numbers = 7! + 18 ´ 6!.
(c) Last digits can only be 25, 50, 75 or
100. But two zeroes are not available 100 is rejected. .
(i) Numbers
finishing with zero, last two digits can be 50 only. For the remaining 6
places, there are 6 choices. So there are 6! such numbers.
(ii) Number
finishing with 5. Second last place can be filled by 2 or 7 only.
No. of choices for
various places is 
\ No. of numbers = 5 ´ 5! ´ 2.
Total number of
numbers = 6! +10 ´ 5!
(d) Number of single-digit numbers = 8
Number of
double-digit numbers = 7 ´
7 = 49
Number of 3-digit
numbers = 7´ 7 ´ 6 = 294
Number of 4-digit
numbers = 7´ 7 ´ 6 ´ 5 = 1470
Number of 5-digit
numbers = 1´ 7 ´ 6 ´ 5 ´
4 = 840
Total number = 2661
(e) A five digit
integers is always greater than 7000. The number of such integer is 5P5
= 5! = 120. For a four digit integer to be greater than 7000, it must begin
with 7, 8 or 9. The number of such integer is (3) (4P3 )
= 72. .
Hence, the required number of
such integers is 120 + 72 = 192.
Example.3
How many 3 letter words can
be formed from the letters of the word CALCUTTA ?
Solution
CALCUTTA has letters CC, AA, TT, U,L. There
are different number of cases where 3 letter words can be formed.
I. 3 alike NOT
POSSIBLE.
II. 2 alike, 1
different.
III. 3
different.
For case (II), the
number of ways it can be done is 3C1 4C1,
since there are 3 pairs of identical letters and 2 other different letters.
Now, these 3 letters can be arranged in 3!/2! ways since 2 of them are
identical. Hence, the total number for case (II) is.
3C1. 4C1.
(3!/2!) = 36. .
For case (III), the
number of ways we can choose 3 different letters from 5 different letters is 5C3.
These 3 letters can be arranged in 3! Ways, so that the number of ways = 5C3.
3!=60.
Hence the total
number of letters that can be formed =36+60 = 96
Example.4
In how may ways 3
boys and 15 girls can sit together in a row such that between any 2 boys at
least 2 girls sit.
Solution
Let x1 girls sit before first
boy, x2 girls between first and second boy, x3 girls
between second and third boy and x4 after the third boy.
Now number of ways in which we can choose
the places for boys is
same as the number
of solution of the equation.
x1 + x2 + x3
+ x4 = 15
Subject to condition x1 ≥ 0, x2 ≥ 2, x3 ≥ 2, x4 ≥ 0
\
Required number of ways = coefficient of x15 in
(1 + x + x2 + ) (x2 +
x3 + ...) (x2 + x3 + ..) (1 + x + x2+...).
= coefficient of x11 in (1 - x)-4
= 4 + 11 - 1C4 - 1 = 14C3
Now 3 boys can be permuted in their places
in 3! ways and 15 girls can be permuted in 15! ways.
So required number of ways = 14C3(3!)(15!)
Example.5
In how may ways can
(2t + 1) identical balls be placed in 3 distinct boxes so that any two boxes
together will contain more balls
than the third.
Solution
The total number of ways to place the balls
disregarding the constraints is 2t+1+3-1C3-1 = 2t+3C2
ways.
The total number of ways to place the balls
so that the first box will have more balls than the other two is t + 3 - 1C3
- 1 = t + 2C2 ways.
(We place (t + 1)
balls in the first box and then divide the rest of t balls in the 3 boxes
arbitrarily).
The same result
applies to the case of 2nd box holding more balls than 1st or 3rd combined and
also for the 3rd box containing more balls than 1st and 2nd combined.
Hence required
number of ways = 2t + 3C2 - 3 t + 2C2
= t/2(t + 1) .
