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The number of ways in which a couple can sit around a table with 6 guests if the couple take consecutive seats is
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!
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
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!
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
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
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
Treat fruits of same kind as identical
The number of proper divisors of 2^{P} x 6^{q} x 15^{r} is
∴ 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
The number of proper divisors of 1800 which are also divisible by 10, is
1800 = 2^{3} x 3^{2} x 5^{2}
∴ 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 number of odd proper divisors of 3^{P} x 6^{m} x 21^{n} is
∴ 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
The number of even proper divisors of 1008 is
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
In a test there were n questions. In the test 2^{n1 }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
The number of students given exactly one wrong answer
2^{n1}  2^{n2}
The number of students giving exactly two wrong answers
= 2^{n2}  2^{n3} etc.
∴ the total number of wrong answers
or
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
Total cases  cases not in favour
The number of ways to distribute 32 different things equally among 4 persons is
Each one will gel 8 objects
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
(The number of ways of dividing in 3 equal groups) x (3!)
= The number of ways to distribute equally among 3 persos.
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
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 of 6digit numbers that can be made with the digits 1, 2. 3 and 4 and having exactly two pairs of digits is
The number will have 2 pairs and 2 different digit.
The number of selections = ^{4}C_{2} x ^{2}C_{2}, and for each selection, number of arrangements =
Therefore, the required number of numbers =
The number of words of four letters containing equal number of vowels and consonants, repetition being allowed, is
The number of selections of 1 pair of vowels and 1 pair of consonants
=^{ 5}C_{1} x ^{21}C_{1}
The number of selections of 2 different vowels and 2 different consonants
=^{ 5}C_{1} x ^{21}C_{2}
∴ the required numbers of words
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
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 = 262
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
The number of arrangements of the letters of the word BHARAT taking 3 at a time is
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
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
In the figure, two 4digit 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 4digit number and in no place the addition is with carrying is
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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