# CAT Permutation and Combination Questions PDF (Set -2)

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This blog contains very important Permutations and Combination Questions and useful tips and tricks for CAT exam.

CAT Permutation and Combination Questions PDF (Set -2):

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Question 1:

Consider the set S = { 1, 2, 3, .., 1000 }. How many arithmetic progressions can be formed from the elements of S that start with l and end with 1000 and have at least 3 elements?

a) 3
b) 4
c) 6
d) 7
e) 8

Question 2:

There are 6 tasks and 6 persons. Task 1 cannot be assigned either to person 1 or to person 2; task 2 must be assigned to either person 3 or person 4. Every person is to be assigned one task. In how many ways can the assignment be done?

a) 144
b) 180
c) 192
d) 360
e) 716

Question 3:

Let S be the set of five-digit numbers formed by the digits 1, 2, 3, 4 and 5, using each digit exactly once such that exactly two odd positions are occupied by odd digits. What is the sum of the digits in the rightmost position of the numbers in S?

a) 228
b) 216
c) 294
d) 192

Question 4:

Three Englishmen and three Frenchmen work for the same company. Each of them knows a secret not known to others. They need to exchange these secrets over person-to-person phone calls so that eventually each person knows all six secrets. None of the Frenchmen knows English, and only one Englishman knows French. What is the minimum number of phone calls needed for the above purpose?

a) 5
b) 10
c) 9
d) 15

Question 5:

Sam has forgotten his friend’s seven-digit telephone number. He remembers the following: the first three digits are either 635 or 674, the number is odd, and the number nine appears once. If Sam were to use a trial and error process to reach his friend, what is the minimum number of trials he has to make before he can be certain to succeed?

a) 10000
b) 2430
c) 3402
d) 3006

Answers and Solutions for CAT Permutation and Combination Questions PDF (Set -2):

Solutions:

The nth term is a + (n-1)d
1000 = 1 + (n-1)d
So, (n-1)d = 999
999 = 3^3 * 37
So, the number of factors is 4*2 = 8
Since there should be at least 3 terms in the series, d cannot be 999.
So, the number of possibilities is 7

If the first task is assigned to either person 3 or person 4, the second task can be assigned in only 1 way. If the first task is assigned to either person 5 or person 6, the second task can be assigned in 2 ways. Therefore, the number of ways in which the first two tasks can be assigned is 2*1 + 2*2 = 6.
The other 4 tasks can be assigned to 4 people in 4! ways.
The total number of ways of assigning the 6 tasks is, therefore, 6*4! = 144.

When the odd numbers occupy places 1 and 3, only 2 or 4 can be in the 5th place. Odd numbers can occupy places 1 and 3 in 3C2*2! = 6 ways. When 2 is at the 5th place, the other odd number and 4 can be arranged in the remaining places in 2 ways. So, 2 occurs at the end 6*2 = 12 times. Similarly, 4 occurs 12 times.
If odd numbers occupy places 1 and 5, then 2 or 4 should come in the 3rd place. The other two numbers can then be arranged in 2 ways in the remaining blanks. So, if 1 is in the first place and 5 is in the 5th place, the other numbers can be arranged in 2*2 = 4 ways. Similar for 1 and 3; 5 and 1; 3 and 1; 5 and 3; 3 and 5. So, 5 occurs 8 times, 1 8 times and 3 8 times. Similar is the case when odd numbers are placed in 3rd and 5th places.
On the whole, 4 occurs 12 times, 2 occurs 12 times, 5, 3 and 1 each occur 16 times. The total is, therefore, 48+24+80+48+16 = 216

Consider there are 6 people numbered 1-3 englishmen and 3-6 frenchmen, let 3 know both english and french.
First call would be between 1-3 then 2-3 such that 3 know secret of all 3 englishmen.
Let 3 call 4 .
Similarly there would be call between 4-5 then 4-6 such that 4 know secret of all 3 frenchmen.
Now 3 would call 4 . Such that 3 and 4 would know secret of all 6 members.
Now to let this know to 1,2,5,6 more 4 calls would be required.
Hence, minimum calls required would be 9.