Permutations and Combinations-Quiz-5

Permutations and Combinations Quiz-5

Dear Readers,

Permutations and Combinations is one of the most important chapters of Algebra in the JEE syllabus and other engineering exams. For JEE Mains, it has 4% weightage and for JEE Advanced, it has 5% weightage..

Q1. There are three copies each of four different books. The number of ways in which they can be arranged in a shelf is

•  12!/(3!)⁴
•  12!/(4!)³
•  21!/((3!)⁴4!)
•  (12!/(4!)³3!)

Solution
Here, we have to divide 12 books into sets 3 books each. Therefore the required number of ways is 12!/((3!)⁴4!) X 4!

Q2.Total number of six-digit numbers that can be formed, having the property that every succeeding digit is greater than the preceding digit, is equal to

•  ⁹C₃
•  ¹⁰C₃
•  ⁹P₃
•  ¹⁰P₃

Solution
x₁ <e; x₂ <e; x₃ <e; x₄ <e; x₅ <e; x₆ <e;, when the number is x₁x₂x₃x₄x₅x₆. Clearly no digit can be zero. Also, all the digits are distinct. So, let us first select six digits from the list of digits which can be done in ⁹C₆ ways. After selecting these digits they can be put only in one order. Thus, total number of such numbers is ⁹C₆ X 1 = ⁹C₃

Q3.  Total number of words that can be formed using all letters of the word ‘BRIJESH’ that neither begins with ‘I’ nor ends with ‘B’ is equal to

•  3720
•  4920
•  3600
•  4800

Solution
Total number of words without any restriction is 7!.
Total number of words beginning with I is 6!.
Total number of words ending with B is 6!.
Total number of words beginning with I and ending with B is 5!.
Thus the total number of required words is 7! - 6! - 6! + 5! = 7! - 2(6!) + 5!

Q4. The number of five-digit numbers that contain 7 exactly once is

•  (41)(9³)
•  (37)(9³)
•  (7)(9⁴)
•  (41)(9⁴)

Solution
If 7 is used at first place, the number of numbers is 9⁴ and for any other four places it is 8 X 9³

Q5.Total number less than 3 X 10⁸ and can be formed using the digits 1, 2, 3 is equal to

•  1/2(3⁹ + 4 X 3⁸)
•  1/2(3⁹ -3)
•  1/2(7 X 3⁸ - 3)
•  1/2(3⁹ - 3 + 3⁸)

Solution

Q6. The number of ways to fill each of the four cells of the table with a distinct natural number such that the sum of the numbers is 10 and the sums of the numbers placed diagonally are equal is

•  4
•  8
•  24
•  6

Solution

Q7.If α = ^mC₂, then ^αC₂ is equal to

•  ^m+1C₄
•  ^m-1C₄
•  3^m+2C₄
•  3^m+1C₄

Solution

Q8.The sum of all the numbers of four different digits that can be made by using the digits 0, 1, 2 and 3 is

•  26664
•  39996
•  38664
•  None of these

Solution
The number of numbers with 0 in the unit’s place is 3! - 2! = 4. Therefore the sum of the digits in the unit’s place is 6 x 0 + 4 x 1 + 4 x 2 + 4 x 3 = 24.
Similarly, for the ten’s and hundred’s places, the number of numbers with 1 or 2 in the thousand’s place is 3! Therefore, the sum of the digits in the thousand’s place is 6 x i + 6 x 2 + 6 x 3 = 36.
Hence, the required sum is 36 x 1000 + 24 x 100 + 24 x 10 + 24

Q9.Among the 8! Permutations of the digits 1,2,3,...,8, consider those arrangements which have the following property. If we take any five consecutive positions, the product of the digits in these positions is divisible by 5. The number of such arrangements is equal to

•  7!
•  2.(7!)
•  ⁷C₄
•  None of these

Solution
Let the arrangement be x₁x₂x₃x₄x₅x₆x₇x₈ and clearly 5 should occupy the position x₄ or x₅. Thus required number is 2(7!)

Q10.20 persons are sitting in a particular arrangement around a circular table. Three persons are to be selected for leaders. The number of ways of selection of three persons such that no two were sitting adjacent to each other is

•  600
•  900
•  698
• None of these

Solution