Shortcut -Permutation and Combination
Dependent and Independent
Events.
Dependent events – Multiplication;
Key word – AND
Independent events - Addition;
Key word – OR
Question:
There 3 busses from city A to city B and there are 5 busses from city B to city C. In how many ways a person can travel from city A to C through B?
Answer: Choosing a bus from city B depends on choosing a bus from city A.
Number of ways = 3 x 5 = 15
Question:
There 3 busses from city A to city B and there are 5 busses from city A to city C. In how many ways a person can travel from city A to C or B?
Answer: Choosing a bus to city B or C are
not dependent.
Number of ways = 3 + 5 = 8
Shortcut -
Permutation and Combination
Permutation – Arrangement with repetition.
$n^{n} $
Question:
In how many ways the letters of the word “ORANGE” can be arranged with repetition?
Answer: n = 6 (n = total
number of elements)
Since all the elements are taken,
Number of arrangements
=$6^{6}$
$n^{r}$
Question:
In how many ways three letters from the word “ORANGE” can be arranged with repetition?
Answer: n =
6;
r = 3 (r = number of elements taken for arrangement)
Number of
arrangements =$ 6^{3}$ = 216
Shortcut -Permutation and Combination
Permutation – Arrangement without repetition. n!
Question:
In how many ways the letters of the word “MANGO” can be arranged without repetition?
Answer: n=5
Since all the elements are taken for
arrangement,
Number of elements = n! = 5! = 120
nPr =$\dfrac{n!}{(n-
r)!}$
Question:
In how many ways any three letters of the word ‘MANGO” can be arranged without repetition?
Answer: n= 5;
r = 3
nPr = 5!/(5 – 3)! = 120/2 = 60 ways
Shortcut -Permutation and
Combination
Permutation – Elements occurring together.
If two elements occur
together
2!(n – 1)!
If three elements occurtogther
3!(n – 2)!
If
four elements occur together
4!(n – 3)! and so on
Question:
In how many ways letters of the word “ORANGE” arranged so that the vowels will always occur together?
Answer: n=6 Three letters ‘O, A
and E’ should occur together.
Since three letters occur together
3! x (6 – 2)! =
6 x 24 = 144
Note:
If there are 4 letters occurring together, the answer is
4!(6 – 3)!
If there are 5 letters occurring together, the answer is
5!(6 – 4)!
Shortcut -Permutation and Combination
Permutation – When
similar kind of elements occur.
$\dfrac{n!}{a! \times b!}$
Question:
In how many ways the letters of the word “ENVIRONMENT” can be arranged?
Answer: n = 11
Let, a = 2 (E is repeated twice)
Let, b = 3 (N is repeated thrice)
Number of arrangements = 11!/(2! x 3!)
=
277200
Question:
In how many ways the letters of the word “PIIZZZAAAA” can be rearranged?
Answer: n = 10
a = 2 (I is
repeated twice)
b = 3 (Z is repeated thrice)
c = 4 (A is repeated four times)
Number of arrangements = 10!/(2! x 3! x 4!)
= 12600
Shortcut -Permutation and Combination
Permutation – Circular arrangement. (n – 1)!
Question:
In how many ways 6 persons can be arranged in a circle?
Answer: n=6
Number of arrangements = (6 –
1)! = 5! = 120
Question:
In how many ways 8 persons can be arranged in a circle?
Answer:n=8
Shortcut -Permutation and Combination
Permutation – Elements occurring together in a circle.
If two elements occur together
2!(n – 2)!
If three elements occur together
3!(n – 3)!
If four
elements occur together
4!(n – 4)!
and so on
Question:
In how many ways 8 persons can be seated around a circular table with two persons always sitting together?
Answer: 2! x (8 – 2)! = 2 x 720 = 1440
Note:
If three persons are sitting together, the answer is
3! x (8 – 3)!
If four
persons are sitting together, the answer is
4! x (8 – 4)!
Shortcut -
Permutation and Combination
Permutation – Arrangement of Necklace.
(n – 1)!/2
Question:
In how many ways a necklace with 8 different colored beads can be arranged?
Answer: n=8
Number of arrangements = (8 –
1)!/2 = 2520
Shortcut -Permutation and Combination
Combination
$nCr =\dfrac{n!}{r!(n-r)!}$
Question:
In how many ways 2 shirts and 3 pants can be selected from 5 shirts and 7 pants?
Answer: Selecting 2 shirts out of 5 = 5C2 = 5!/[2!(5 –
2)!] = 10
Selecting 3 shirts out of 7 = 7C3 = 7!/[3!(7 – 3)!] = 35
Since
the two events are dependent,
Total ways of selecting = 10 x 35 = 350
Question:
In how many ways 2 shirts or 3 pants can be selected from 5 shirts and 7 pants?
Answer: Selecting 2 shirts out of 5 = 5C2 = 5!/[2!(5 – 2)!]
= 10
Selecting 3 shirts out of 7 = 7C3 = 7!/[3!(7 – 3)!] = 35
Since the two
events are independent,
Total ways of selecting = 10 + 35 = 45