Test:- Permutations And Combinations - 12

# Test:- Permutations And Combinations - 12

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## 20 Questions MCQ Test Topic-wise Tests & Solved Examples for IIT JAM Mathematics | Test:- Permutations And Combinations - 12

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Test:- Permutations And Combinations - 12 - Question 1

### The number of ways in which a couple can sit around a table with 6 guests if the couple take consecutive seats is

Detailed Solution for Test:- Permutations And Combinations - 12 - Question 1

A couple and 6 guests can be arranged in (7 - 1)! ways. But the two people forming the couple can be arranged among themselves in 2! ways.
∴ the required number of ways = 6! x 2!

Test:- Permutations And Combinations - 12 - Question 2

### The number of ways in which 20 different pearls of two colours can be set alternately on a necklace, there being 10 pearls of each colour, is

Detailed Solution for Test:- Permutations And Combinations - 12 - Question 2

Ten pearls of one colour can be arranged in 1/2 x (10 - 1)! ways. The number of arrangements of 10 pearls of the other colour in 10 places between the pearls of the first colour = 10!
∴ the required number of ways = 1/2 x 9! x 10!

Test:- Permutations And Combinations - 12 - Question 3

### If r > p > q, the number of different selections of p + q things taking r at a time, where p things are identical and q other things are identical, is

Detailed Solution for Test:- Permutations And Combinations - 12 - Question 3

The number of selections of p things from p identical things and r - p things from q identical things = 1 x 1 Similarly in all other cases,
∴ the total number of ways
= p - (r - q )+ 1 or q - (r - p) + 1
= p + q - r + l

Test:- Permutations And Combinations - 12 - Question 4

There are 4 mangoes, 3 apples. 2 oranges and 1 each of 3 other varieties of fruits. The number of ways of selecting atleast one fruit of each kind is

Detailed Solution for Test:- Permutations And Combinations - 12 - Question 4

Treat fruits of same kind as identical

Test:- Permutations And Combinations - 12 - Question 5

The number of proper divisors of 2P x 6q x 15r is

Detailed Solution for Test:- Permutations And Combinations - 12 - Question 5 ∴ the number of proper divisors
= {total number of selections from (p + q)
twos, (q+r) threes and r five} - 2
= (p + q + 1)(q + r + 1)(r + 1)-2

Test:- Permutations And Combinations - 12 - Question 6

The number of proper divisors of 1800 which are also divisible by 10, is

Detailed Solution for Test:- Permutations And Combinations - 12 - Question 6

1800 = 23 x 32 x 52
∴ the required number of proper divisors
= total number of selections of at least one 2 and one 5 from 2,2,2,3,3,5,5
= 3 x (2 + 1) x 2 =18

Test:- Permutations And Combinations - 12 - Question 7

The number of odd proper divisors of 3P x 6m x 21n is

Detailed Solution for Test:- Permutations And Combinations - 12 - Question 7 ∴ the required number of proper divisors
= total number of selections of zero 2 and any number of 3’s or 7’s.
= (p + m + n + 1) (n + 1) - 1

Test:- Permutations And Combinations - 12 - Question 8

The number of even proper divisors of 1008 is

Detailed Solution for Test:- Permutations And Combinations - 12 - Question 8

1008 = 24 x 32 x 7
∴ the required number of even proper divisors
= total number of selections of at least one 2 and any number of 3's or 7's
= (4 x (2+1) x (1+1)-1

Test:- Permutations And Combinations - 12 - Question 9

In a test there were n questions. In the test 2n-1 students gave wrong answers to i questions where i = 1, 2, 3, ..., n. If the total number of wrong answers given is 2047 then n is

Detailed Solution for Test:- Permutations And Combinations - 12 - Question 9

The number of students given exactly one wrong answer
2n-1 - 2n-2
The number of students giving exactly two wrong answers
= 2n-2 - 2n-3 etc.
∴ the total number of wrong answers or Test:- Permutations And Combinations - 12 - Question 10

The number of ways to give 16 different things to three persons A, 13, C so that B gets 1 more than A and C gets 2 more than B, is

Detailed Solution for Test:- Permutations And Combinations - 12 - Question 10

Total cases - cases not in favour

Test:- Permutations And Combinations - 12 - Question 11

The number of ways to distribute 32 different things equally among 4 persons is

Detailed Solution for Test:- Permutations And Combinations - 12 - Question 11

Each one will gel 8 objects

Test:- Permutations And Combinations - 12 - Question 12

If 3n different things can be equally distributed among 3 persons in k ways then the number of ways to divide the 3n things in 3 equal groups is

Detailed Solution for Test:- Permutations And Combinations - 12 - Question 12

(The number of ways of dividing in 3 equal groups) x (3!)
= The number of ways to distribute equally among 3 persos.

Test:- Permutations And Combinations - 12 - Question 13

In a packet there are rn different books, n different pens and p different pencils. The number of selections of atleast one article of each type from the packet is

Detailed Solution for Test:- Permutations And Combinations - 12 - Question 13

The required number of ways
= Total number of ways of selecting any number of books from m different books, any number of pens from n different, pens and any number of pencils from p different pencils - 1. Test:- Permutations And Combinations - 12 - Question 14

The number of 6-digit numbers that can be made with the digits 1, 2. 3 and 4 and having exactly two pairs of digits is

Detailed Solution for Test:- Permutations And Combinations - 12 - Question 14

The number will have 2 pairs and 2 different digit.
The number of selections = 4C2 x 2C2, and for each selection, number of arrangements = Therefore, the required number of numbers = Test:- Permutations And Combinations - 12 - Question 15

The number of words of four letters containing equal number of vowels and consonants, repetition being allowed, is

Detailed Solution for Test:- Permutations And Combinations - 12 - Question 15

The number of selections of 1 pair of vowels and 1 pair of consonants
= 5C1 x 21C1
The  number of selections of 2 different vowels and 2 different consonants
= 5C1 x 21C2
∴ the required numbers of words Test:- Permutations And Combinations - 12 - Question 16

The number of ways in which 6 different balls can be put in two boxes of different sizes so that no box remains empty is

Detailed Solution for Test:- Permutations And Combinations - 12 - Question 16

Each ball can be put in 2 ways (either in one box or the other)
∴ 6 balls can be put in 2 x 2 x ... to six times, i.e.,26 ways. But in two of the ways one box is empty. So, the required number of ways = 26-2

Test:- Permutations And Combinations - 12 - Question 17

A shopkeeper sells three varieties of perfumes and he has a large; number of bottles of the same size of each variety' in his stock. There are 5 places in a row in his showcase. The number of different ways of displaying the three varrieties of perfumes in the showcase is

Detailed Solution for Test:- Permutations And Combinations - 12 - Question 17 Test:- Permutations And Combinations - 12 - Question 18

The number of arrangements of the letters of the word BHARAT taking 3 at a time is

Detailed Solution for Test:- Permutations And Combinations - 12 - Question 18 Test:- Permutations And Combinations - 12 - Question 19

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 Detailed Solution for Test:- Permutations And Combinations - 12 - Question 19

The natural numbers are 1, 2, 3, 4. Clearly, in one diagonal we have to place 1, 4 and in the other 2, 3.
The number of ways in (i) - 2! * 2! = 4
the number of ways in (ii) = 2! x 2! = 4
∴ the total number of ways = 8

Test:- Permutations And Combinations - 12 - Question 20

In the figure, two 4-digit numbers are to be formed by filling the places with digits. The number of different ways in which the places can be filled by digits so that the sum of the numbers formed is also a 4-digit number and in no place the addition is with carrying is Detailed Solution for Test:- Permutations And Combinations - 12 - Question 20

If 0 is placed in the units place of the upper number then the units pice of the lower number can be filled in 10 ways (Filling by ay one of 0, 1, 2......9).
If 1 placed in the units placed of the upper number then the unit place of the lower number can be filled in 9 ways (filling by any one of 0, 1, 2, ..., 8), etc.
the units column can be filled in 10 - 9 + 8 + ... * 1, i.e., 55 ways. Similarly for the second and the third column. The number of ways for the fourth column = 8 + 7 + ... + 1 = 36
∴ the required number of ways = 55 x 55 x 55 x 36

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