CONTENTS:
1 |
Definitions |
2 |
Formulas |
3 |
35 Selected Problems |
4 |
Solutions to the Selected Problems |
Definitions
Permutation
A permutation is a mathematical technique that
determines the number of possible arrangements in a set when the order of the
arrangements matters. Common mathematical problems involve choosing only
several items from a set of items in a certain order.
Combination
In mathematics, a combination refers to the
technique that determines the number of possible selections of items / things
from a set where the order of selections does not matter. It’s a way of
choosing a group of objects / things without regard to the arrangement or
sequence in which they are picked.
Formulas
1 |
Permutation
of ‘n’ different things taken ‘r’ at a time (nPr): |
|
nPr = n! / (n – r)! |
|
|
2 |
Permutation
of ‘n’ different things taken all at a time (nPn): |
|
nPn = nP(n – 1) = n! |
|
|
3 |
0!
= 1 |
|
|
4 |
1!
= 1 |
|
|
5 |
n!
= n × (n – 1) × (n – 2) ×..........× 1. [‘n!’
is read as ‘n factorial’ ] |
|
|
6 |
nP0 = 1 |
|
|
7 |
nP1 = n |
|
|
8 |
nPr = n × (n – 1)P(r – 1) |
|
|
9 |
nPr = (n – 1)Pr + r
× (n – 1)P(r – 1) |
|
|
10 |
Permutation
of ‘n’ things taken all at a time, where ‘n’ is represented by ‘p’ equal
things, ‘q’ equal things and ‘r’ equal things |
|
=
n! / (p! × q! × r!) |
|
|
11 |
Permutation
of ‘n’ different things each of which can be repeated ‘r’ number of times
(For example, In
how many ways ‘r’ letters can be posted in ‘n’ different boxes?) |
|
=
nr |
|
|
12 |
Number
of ways in which ‘n’ persons can be seated at a round table with respect to
the table |
|
=
n! |
|
|
13 |
Number
of ways in which ‘n’ persons can be seated at a round table with respect to
each other |
|
=
(n – 1)! |
|
|
14 |
Number
of ways in which ‘n’ different stones/beads can be placed in a necklace |
|
=
½ × (n – 1)! |
|
|
15 |
Permutation
of ‘n’ different things taken ‘r’ at a time in which ‘p’ particular things
never occur |
|
=
(n – p)Pr |
|
|
16 |
Permutation
of ‘n’ different things taken ‘r’ at a time in which ‘p’ particular things
occupy stated places |
|
=
(n – p)P(r – p) |
|
|
17 |
Permutation
of ‘n’ different things taken ‘r’ at a time in which ‘p’ particular things
are always present |
|
=
(n – p)P(r – p) × rPp |
|
|
18 |
Permutation
of ‘n’ different things taken ‘r’ at a time in which ‘p’ particular things
shall always occur together in an assigned order |
|
=
(n – p)P(r – p) × (r – p + 1) |
|
|
19 |
Permutation
of ‘n’ different things taken all at a time in which ‘p’ particular things
shall always occur in an assigned order |
|
=
n! /p! |
|
|
1 |
Combination
of ‘n’ different things taken ‘r’ at a time (nCr): |
|
nCr = n! ÷ [r! × (n – r)!] |
|
|
2 |
nCr = nPr ÷ r! |
|
|
3 |
nCr = nC(n – r) |
|
|
4 |
nCr = (n – 1)Cr + (n
– 1)C(r – 1) |
|
|
5 |
nC0 = 1 |
|
|
6 |
nC1 = n |
|
|
7 |
nCn = 1 |
|
|
8 |
If
nCr = nCs, either r = s, or r + s
= n |
|
|
9 |
nCr ÷ nC(r – 1) = (n
– r + 1) ÷ r |
|
|
10 |
nCr + nC(r – 1) = (n
+ 1)Cr |
|
|
11 |
Number
of combination of ‘n’ different things taken ‘r’ at a time where ‘p’ things
will always occur |
|
=
(n – p)C(r – p) |
|
|
12 |
Number
of combination of ‘n’ different things taken ‘r’ at a time where ‘p’ things
will never occur |
|
=
(n – p)Cr |
|
|
13 |
(m
+ n)C(p + q)
= mCp + nCq |
|
|
14 |
nC1 + nC2 + nC3
+........+ nCn = 2n − 1 |
|
|
15 |
nC0 + nC1 + nC2
+ nC3 +........+ nCn = 2n |
|
|
16 |
Combination
of ‘n’ things taken all together where ‘n’ is represented by ‘p’ equal
things, ‘q’ equal things and ‘r’ equal things |
|
=
(p + 1)(q + 1)(r + 1) − 1 |
|
|
17 |
Number
of ways in which two groups out of (m + n) things (m ≠ n) can be formed so
that one group consists of ‘m’ things and the other group consists of ‘n’
things |
|
=
(m + n)Cm = (m + n)! ÷ (m! × n!) |
|
|
18 |
Number
of ways in which ‘m’ things can be selected from (m + n) things (m ≠ n) |
|
=
(m + n)Cm = (m + n)! ÷ (m! × n!) |
|
|
19 |
Number
of ways in which two groups out of (m + n) things (m = n) can be formed so
that each group consists of equal number of things |
|
=
(1/2!) × [(2m)! ÷ 2(m!)] |
|
|
20 |
Number
of ways in which three groups out of (m + n + p) things (m ≠ n ≠ p) can be
formed so that the groups contain ‘m’ things, ‘n’ things and ‘p’ things
respectively |
|
=
(m+n+p)Cm × (n+p)Cn × pCp |
|
=
(m+n+p)! ÷ (m! × n! × p!) |
|
|
21 |
Number
of ways in which three groups out of (m + n + p) things (m = n = p) can be
formed so that each group contains equal number of things |
|
=
(1/3!) × [(3m)! ÷ 3(m!)] |
|
|
22 |
Number
of ways in which ‘2m’ things can be divided equally amongst 2 persons |
|
=
(2m)! ÷ 2(m!) |
|
|
23 |
Number
of ways in which ‘3m’ things can be divided equally amongst 3 persons |
|
=
(3m)! ÷ 3(m!) |
|
|
CMA
– Foundation
Permutation
and Combination
Selected
Problems
1. In how many ways
can 5 candidates be screened for 3 vacancies for 3 different job-profiles?
2. In how many ways
3 cheques can be given to 4 employees when each employee is eligible to receive
any number of cheques?
3. Find the number
of arrangements which could be made with the letters of the word “APPLE”.
4. In how many ways
8 boys can form a ring?
5. In how many ways
8 different beads can be placed in a necklace?
6. If 1/ (4!) + 1/
(5!) = x/ (6!), find ‘x’.
7. There are 21
boys and 19 girls in a class. In how many ways can one boy and one girl be
selected?
8. In how many ways
the 1st and 2nd prizes of maths and stats can be given in
a class of 40 students?
9. How many words
can be formed by using the letters of the word “ALLAHABAD”, so that vowels will
occupy even places?
10. There are 5
doors and 4 windows in a room. In how many ways a cat can enter and leave that
room?
11. In how many
ways 3 rows can be selected for vaccination out of 5 rows of students of a
college assuming vaccination can be done only for one row at a time?
12. If (n + 1)P3:
nP2 = 27: 9, find ‘n’.
13. If 8 × (n
– 1)P3 = (n – 1)P4, find ‘n’.
14. If nP5
= 42 × nP3, find ‘n’.
15. In how many
ways 6 mobiles can be gifted to 6 employees, if no employee is eligible to
receive more than one mobile?
16. In how many
ways can 8 articles numbered from 1 to 8 be arranged, if even numbered articles
are same?
17. In how many
ways 6 books out of 10 different books can be arranged in a bookshelf so that 3
particular books are always together?
18. In how many
ways can the letters of the word “TABLE” be arranged so that the vowels are
always - (i) together, (ii) separated?
19. Find in how
many ways the letters of the word “PURPOSE” can be rearranged
i)
Keeping the position of the vowels fixed;
ii)
Without changing the relative positions of the vowels
and consonants.
20. How many numbers between 5,000 and 6,000 can be
formed with the digits 3, 4, 5, 6, 7, 8 when none of the digits will be
repeated in any of the numbers so formed?
21. In how many ways can the letters of the word
SUNDAY be arranged? How many of them do not begin with ‘S’? How many of them do
not begin with ‘S’, but end with ‘Y’?
22. In how many ways 5 boys and 5 girls can take their
seats out of 10 seats in a round table, so that no two girls will sit side by
side?
23. From a group of 15 men, how many selections of 9
men can be made so as to exclude 3 particular men?
24. There are 7 candidates for a post. In how many
ways can a selection of 4 be made amongst them, so that:
(a)
2 persons whose qualifications are below par are
excluded?
(b)
2 persons with good qualifications are included?
25. In how many ways can a committee of 3 ladies and 4
gentlemen be appointed from a meeting consisting of 8 ladies and 7 gentlemen?
What will be the number of ways if Mrs. Sen refuses to serve in a committee
having Mr. Sharma as a member?
26. From 7 gentlemen and 4 ladies a committee of 5 is
to be formed. In how many ways can this be done to include at least one lady?
27. In how many ways can a boy invite one or more of
his 5 friends?
28. In how many ways 15 things can be divided into
three groups of 4, 5 and 6 things respectively?
29. In how many ways can the letters of the word
SIGNAL be arranged so that the vowels occupy only the odd places?
30. How many triangles can be made by joining the
vertices of a decagon? How many diagonals will it have?
31. There are 10 persons in a party. If one can shake
hand with others, how many handshakes can be made?
32. How many diagonals are there in a hexagon?
33. There are 12 points in a plane of which 5 are
collinear. Find the number of triangles that can be formed with these points.
34. In how many ways can a team of 11 be chosen from
14 football players two of whom can only be goalkeeper?
35. A polygon has 27 diagonals. Find number of its
sides.
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