# Fundamentals for permutation and combination problems

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Fundamentals for permutation and combination problems

If there are n events and if the first event can occur in m1 ways, the second event can occur in m2 ways after the first event has occurred, the third event can occur in m3 ways after the second event has occurred and so on, then all the n events can occur in:

m1*m2*m3*…*mn−1*mn ways.

The number of permutations of n objects taken all at a time is n!

nPr =  n!/ (n-r)!

nCr =  n!/r!(n-r)!

nCr = nC(n-r)
nCr =  nPr/r!

0!=1
1!=1
nPn=n!
nP1=n
nC1=n
nCn=1

There are 5 freshmen, 8 sophomores, and 7 juniors in a chess club. A group of 6 students will be chosen to compete in a competition.

1. How many combinations of students are possible if the group is to consist of exactly 3 freshmen?
A.5000
B.4550
C.4000
D.3550
Solution:
Here we need the number of possible combinations of 3 out of 5 freshmen,5C3, and the number of possible combinations of 3 out of the 15 sophomores and juniors,15. C3
Note that we want 3 freshmen and 3 students from the other classes.
Therefore, we multiply the number of possible groups of 3 of the 5 freshmen times the number of possible groups of 3 of the 15 students from the other classes.
5C3×15. C3= 4,550

2. How many combinations of students are possible if the group is to consist of exactly 3 freshmen and 3 sophomores?
A.360
B.460
C.560
D.660
Solution:
This second part of the problem is similar to the first except that the choice of the second group of 3 comes only from the sophomores.
Again we want 3 freshmen and 3 sophomores so we multiply the number of groups of freshmen times the number of groups of sophomores.
5C3×8C3= 560

3 How many ways can 4 prizes be given away to 3 boys, if each boy is eligible for all the prizes?
A.256
B.24
C.12
D.None of these
Solution:
Let the 3 boys be B1, B2, B3 and 4 prizes be P1, P2, P3 and P4.
Now B1 is eligible to receive any of the 4 available prizes (so 4 ways)
B2 will receive prize from rest 3 available prizes(so 3 ways)
B3 will receive his prize from the rest 2 prizes available(so 2 ways)

Finally B4 will receive the the last remaining prize (so 1 way)
So total ways would be: 4*3*2*1=24 Ways

Hence, the 4 prizes can be distributed in 24 ways.

4 Find the number of ways in which 8064 can be resolved as the product of two factors?
A.22
B.24
C.21
D.20
Solution:
Total number of ways in which 8064 can be resolved as the product of two factors is 24 as below:
(1,8064), (2,4032), (3,2688), (4,2016), (6,1344), (7,1152),(8,1008), (9,896), (12,672), (14,576), (16,504), (18,448), (21,884), (24,336), (28,288), (32,252), (36,224), (42,192), (48,168), (56,144), (63,128), (68,126), (72,112), (84,96).

5 The letter of the word LABOUR are permuted in all possible ways and the words thus formed are arranged as in a dictionary. What is the rank of the word LABOUR?
A.275
B.251
C.240
D.242
Solution:
The order of each letter in the dictionary is ABLORU.
Now, with A in the beginning, the remaining letters can be permuted in 5! ways.
Similarly, with B in the beginning, the remaining letters can be permuted in 5! ways.
With L in the beginning, the first word will be LABORU, the second will be LABOUR.
Hence, the rank of the word LABOUR is 5!+5!+2= 242

6 A class photograph has to be taken. The front row consists of 6 girls who are sitting. 20 boys are standing behind. The two corner positions are reserved for the 2 tallest boys. In how many ways can the students be arranged?
A.6! × 1440
B.18! × 1440
C.18! ×2! × 1440
D.None of these

Solution:
Two tallest boys can be arranged in 2! ways, rest 18 can be arranged in 18! ways.
Girls can be arranged in 6! ways.
Total number of ways in which all the students can be arranged =2!×18!×6!= 18! ×1440

7 Ten different letters of alphabet are given, words with 5 letters are formed from these given letters. Then, the number of words which have at least one letter repeated is:
A.69760
B.30240
C.99748
D.42386
Solution:
Number of words which have at least one letter replaced:
= Total number of words – total number of words in which no letter is repeated
⇒105–P516
⇒100000−30240= 69760

8 12 chairs are arranged in a row and are numbered 1 to 12. 4 men have to be seated in these chairs so that the chairs numbered 1 to 8 should be occupied and no two men occupy adjacent chairs. Find the number of ways the task can be done.
A.360
B.384
C.432
D.470
Solution:
Given there are 12 numbered chairs, such that chairs numbered 1 to 8 should be occupied.
_X_ , 2,3,4,5,6,7,_X_,9,10,11,12.

The various combinations of chairs that ensure that no two men are sitting together are listed.
(1,3,5,__). The fourth chair can be 5,6,10,11 or 12, hence 5 ways.
(1,4,8,__), The fourth chair can be 6,10,11 or 12 hence 4 ways.
(1,5,8,__), the fourth chair can be 10,11 or 12 hence 3 ways.
(1,6,8,__), the fourth chair can be 10,11 or 12 hence 3 ways.
(1,8,10,12) is also one of the combinations.

Hence, 16 such combinations exist.
In case of each these combinations we can make the four men inter arrange in 4! ways.
Hence, the required result =16×4!= 384

9 In how many ways can six different rings be worn on four fingers of one hand?
A.10
B.12
C.15
D.16