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!)
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
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
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)
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)
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
Q7.If α = mC2, then αC2 is equal to
  •  m+1C4
  •  m-1C4
  •  3m+2C4
  •  3m+1C4
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
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
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

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Permutations and Combinations-Quiz-5
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