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Vizag Steel Plant RINL model questions for permutation and combination to

practice.Permutation and combination fundementals theroy,keynotes,practice

questions and mock on line testsFundamentals for permutation and combination problems

If there are n events and if the first event can occur in m1ways, the second event can occurin m2ways 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*...*mn1*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!=11!=1nPn=n!nP1=nnC1=nnCn=1

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

1. How many combinations of students are possible if the group is to consist of exactly 3freshmen?A.5000

B.4550C.4000D.3550

Answer

(B)Solution:Here we need the number of possible combinations of 3 out of 5 freshmen,5C3, andthe number of possible combinations of 3 out of the 15 sophomores and

juniors,15.C3Note 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.460C.560

D.660Answer (C)

Solution:

This second part of the problem is similar to the first except that the choice of thesecond 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.

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5C38C3= 560

3 How many ways can 4 prizes be given away to 3 boys, if each boy is eligible for all theprizes?

A.256B.24

C.12D.None of these

Answer (B)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.22B.24C.21

D.20

Answer (B)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.275B.251

C.240

D.242Answer (D)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. 20boys are standing behind. The two corner positions are reserved for the 2 tallest boys. Inhow many ways can the students be arranged?

A.6! 1440

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B.18! 1440

C.18! 2! 1440

D.None of theseAnswer (B)

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 thesegiven letters. Then, the number of words which have at least one letter repeated is:

A.69760

B.30240C.99748D.42386

Answer (A)

Solution:

Number of words which have at least one letter replaced:= Total number of words - total number of words in which no letter is repeated105P51610000030240= 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 occupyadjacent chairs. Find the number of ways the task can be done.A.360

B.384C.432D.470

Answer

(B)Solution:Given there are 12 numbered chairs, such that chairs numbered 1 to 8 should beoccupied.

_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 =164!= 384

9 In how many ways can six different rings be worn on four fingers of one hand?A.10

B.12

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C.15

D.16

Answer (C)Solution:

Required number of ways =6C4=65/2= 15 ways.

10 There are 12 yes or no questions. How many ways can these be answered?

A.4096B.2048

C.1024

D.144Answer (A)

Solution:

Each of the questions can be answered in 2 ways (yes or no)Therefore, no. of ways of answering 12 questions =212= 4096 ways.