Solutions to questions on permutations and combinations

Have given below solutions to the questions on permutations and combinations

1. a, b, c are three distinct integers from 2 to 10 (both inclusive). Exactly one of ab, bc and ca is odd. abc is a multiple of 4. The arithmetic mean of a and b is an integer and so is the arithmetic mean of a, b and c. How many such triplets are possible (unordered triplets)

Exactly one of ab, bc and ca is odd => Two are odd and one is even

abc is a multiple of 4 => the even number is a multiple of 4

The arithmetic mean of a and b is an integer => a and b are odd

and so is the arithmetic mean of a, b and c. => a+ b + c is a multiple of 3

c can be 4 or 8.
c = 4; a, b can be 3, 5 or 5, 9
c = 8; a, b can be 3, 7 or 7, 9

Four triplets are possible

2. A seven-digit number comprises of only 2's and 3's. How many of these are multiples of 12?

Number should be a multiple of 3 and 4. So, the sum of the digits should be a multiple of 3. WE can either have all seven digits as 3, or have three 2's and four 3's, or six 2's and a 3. (The number of 2's should be a multiple of 3).

For the number to be a multiple of 4, the last 2 digits should be 32. Now, let us combine these two.

All seven 3's - No possibility
Three 2's and four 3's - The first 5 digits should have two 2's and three 3's in some order. No of possibilities = 5!/3!*2! = 10

Six 2's and one 3 - The first 5 digits should all be 2's. So, there is only one number 2222232

3. How many factors of 2^5 * 3^6 * 5^2 are perfect squares?

Any factor of this number should be of the form 2^a * 3^b * 5^c. For the factor to be a perfect square a,b,c have to be even. a can take values 0, 2, 4. b can take values 0,2, 4, 6 and c can take values 0,2. Total number of perfect squares = 3 * 4 * 2 = 24


4. How many factors of 2^4 * 5^3 * 7^4 are odd numbers?

Any factor of this number should be of the form 2^a * 3^b * 5^c. For the factor to be an odd number, a should be 0. b can take values 0,1, 2,3, and c can take values 0, 1, 2,3, 4. Total number of odd factors = 4 * 5 = 20

5. A number when divided by 18 leaves a remainder 7. The same number when divided by 12 leaves a remainder n. How many values can n take?

Number can be 7, 25, 43, 61, 79.
Remainders when divided by 12 are 7 and 1.

n can take exactly 2 values

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IIM CAT Preparation Tips: Solutions to questions on permutations and combinations

Aug 21, 2011

Solutions to questions on permutations and combinations

Have given below solutions to the questions on permutations and combinations

1. a, b, c are three distinct integers from 2 to 10 (both inclusive). Exactly one of ab, bc and ca is odd. abc is a multiple of 4. The arithmetic mean of a and b is an integer and so is the arithmetic mean of a, b and c. How many such triplets are possible (unordered triplets)

Exactly one of ab, bc and ca is odd => Two are odd and one is even

abc is a multiple of 4 => the even number is a multiple of 4

The arithmetic mean of a and b is an integer => a and b are odd

and so is the arithmetic mean of a, b and c. => a+ b + c is a multiple of 3

c can be 4 or 8.
c = 4; a, b can be 3, 5 or 5, 9
c = 8; a, b can be 3, 7 or 7, 9

Four triplets are possible

2. A seven-digit number comprises of only 2's and 3's. How many of these are multiples of 12?

Number should be a multiple of 3 and 4. So, the sum of the digits should be a multiple of 3. WE can either have all seven digits as 3, or have three 2's and four 3's, or six 2's and a 3. (The number of 2's should be a multiple of 3).

For the number to be a multiple of 4, the last 2 digits should be 32. Now, let us combine these two.

All seven 3's - No possibility
Three 2's and four 3's - The first 5 digits should have two 2's and three 3's in some order. No of possibilities = 5!/3!*2! = 10

Six 2's and one 3 - The first 5 digits should all be 2's. So, there is only one number 2222232

3. How many factors of 2^5 * 3^6 * 5^2 are perfect squares?

Any factor of this number should be of the form 2^a * 3^b * 5^c. For the factor to be a perfect square a,b,c have to be even. a can take values 0, 2, 4. b can take values 0,2, 4, 6 and c can take values 0,2. Total number of perfect squares = 3 * 4 * 2 = 24


4. How many factors of 2^4 * 5^3 * 7^4 are odd numbers?

Any factor of this number should be of the form 2^a * 3^b * 5^c. For the factor to be an odd number, a should be 0. b can take values 0,1, 2,3, and c can take values 0, 1, 2,3, 4. Total number of odd factors = 4 * 5 = 20

5. A number when divided by 18 leaves a remainder 7. The same number when divided by 12 leaves a remainder n. How many values can n take?

Number can be 7, 25, 43, 61, 79.
Remainders when divided by 12 are 7 and 1.

n can take exactly 2 values

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1 Comments:

At 12:51 PM , Blogger kipkirui Langat said...

i have not understood clearly

 

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