Test:- Permutations And Combinations - 12 - Mathematics MCQ

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!

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!

 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

Treat fruits of same kind as identical

∴ 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

 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

∴ 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

 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

 The number of students given exactly one wrong answer
2^n-1 - 2^n-2
The number of students giving exactly two wrong answers
= 2^n-2 - 2^n-3 etc.
∴ the total number of wrong answers 

or 

Total cases - cases not in favour

Each one will gel 8 objects

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

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.

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 =  

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

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

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

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