12.
A five-digit numbers divisible by 3 is to be formed using the numerals 0, 1, 2, 3, 4 and 5, without repetition. The total number of ways this can be done is
KEY CONCEPT : We know that a number is divisible by 3 if the sum of its digits is divisibly by 3.
Now out of 0, 1, 2, 3, 4, 5 if we take 1, 2, 3, 4, 5 or 0 , 1, 2, 4, 5 then the 5 digit numbers will be divisible by 3. Case I : Number of 5 digit numbers formed using the digits 1, 2, 3, 4, 5 = 5! = 120 Case II : Taking 0, 1, 2, 4, 5 if we make 5 digit number then
I place can be filled in = 4 ways (0 can not come at I place)
II place can be filled in = 4 ways
III place can be filled in = 3 ways
IV place can be filled in = 2 ways
V place can be filled in = 1 ways
∴ Total numbers are $$ = 4 \times 4! = 96$$
Thus total numbers divisible by 3 are = 120 + 96 = 216
13.
The total number of integral solutions for $$\left( {x,y,z} \right)$$ such that $$xyz = 24$$ is
$$24 = 2 \cdot 3 \cdot 4,2 \cdot 2 \cdot 6,1 \cdot 6 \cdot 4,1 \cdot 3 \cdot 8,1 \cdot 2 \cdot 12,1 \cdot 1 \cdot 24$$
(as product of three positive integers)
∴ the total number of positive integral solutions of $$xyz = 24$$ is equal to $$3!\, + \frac{{3!}}{{2!}} + 3!\, + 3!\, + 3!\, + \frac{{3!}}{{2!}},{\text{i}}{\text{.e}}{\text{., }}30.$$
Any two of the factors in each factorization may be negative.
∴ the number of ways to associate negative sign in each case is $$^3{C_2},$$ i.e., 3.
∴ the total number of integral solutions $$ = 30 + 3 \times 30 = 120.$$
14.
There are three men and seven women taking a dance class. Number of different ways in which each man is paired with a woman partner, and the four remaining women are paired into two pairs each of two is
$$10 < _{7w}^{3m}3$$ women can be selected in $$^7{C_3}\,$$ ways and can be paired with 3 men in $$3!$$ ways. Remaining 4 women can be grouped into two couples in $$\frac{{4!}}{{\left( {2! \cdot 2! \cdot 2!} \right)}} = 3.$$
Therefore, total $$ = {\,^7}{C_3} \cdot 3! \cdot 3 = 630.$$
15.
The number of odd proper divisors of $${3^p} \cdot {6^m} \cdot {21^n}$$ is
$${3^p} \cdot {6^m} \cdot {21^n} = {2^m} \cdot {3^{p + m + n}} \cdot {7^n}$$
∴ the required number of proper divisors
= total number of selections of zero 2 and any number of $$3’s$$ and $$7’s$$
$$ = \left( {p + m + n + 1} \right)\left( {n + 1} \right) - 1.$$
16.
A teaparty is arranged for 16 people along two sides of a large table with 8 chairs on each side. Four men want to sit on one particular side and two on the other side. The number of ways in which they can be seated is
There are 8 chairs on each side of the table. Let the sides be represented by $$A$$ and $$B.$$ Let four persons sit on side $$A,$$ then number of ways of arranging 4 persons on 8 chairs on side $$A = {\,^8}{P_4}$$ and then two persons sit on side $$B.$$ The number of ways of arranging 2 persons on 8 chairs on side $$B = {\,^8}{P_2}$$ and the remaining 10
persons can be arranged in remaining 10 chairs in $$10!$$ ways.
Hence, the total number of ways in which the persons can be arranged
$$ = {\,^8}{P_4} \times {\,^8}{P_2} \times 10! = \frac{{8!8!10!}}{{4!6!}}$$
17.
$$ABCD$$ is a convex quadrilateral. 3, 4, 5 and 6 points are marked on the
sides $$AB, BC, CD$$ and $$DA$$ respectively. The number of triangles with vertices on different sides is
The number of triangles with vertices on sides $$AB,BC,CD = {\,^3}{C_1} \times {\,^4}{C_1} \times {\,^5}{C_1}.$$
Similarly for other cases.
∴ the total number of triangles
$$ = {\,^3}{C_1} \times {\,^4}{C_1} \times {\,^5}{C_1} + {\,^3}{C_1} \times {\,^4}{C_1} \times {\,^6}{C_1} + {\,^3}{C_1} \times {\,^5}{C_1} \times {\,^6}{C_1} + {\,^4}{C_1} \times {\,^5}{C_1} \times {\,^6}{C_1} = 342.$$
18.
The set $$S = \left\{ {1,2,3,......,12} \right\}$$ is to be partitioned into three sets $$A, B, C$$ of equal size. Thus $$A \cup B \cup C = S,$$ $$A \cap B = B \cap C = A \cap C = \phi .$$ The number of ways to partition $$S$$ is
$$\eqalign{
& {\text{Set }}S = \left\{ {1,2,3,......,12} \right\} \cr
& A \cup B \cup C = S,A \cap B = B \cap C = A \cap C = \phi . \cr} $$
∴ The number of ways to partition
$$ = {\,^{12}}{C_4} \times {\,^8}{C_4} \times {\,^4}{C_4} = \frac{{12!}}{{4!8!}} \times \frac{{8!}}{{4!4!}} \times \frac{{4!}}{{4!0!}} = \frac{{12!}}{{{{\left( {4!} \right)}^3}}}$$
19.
Let $$1 \leqslant m < n \leqslant p.$$ The number of subsets of the set $$A = \left\{ {1,2,3,.....,p} \right\}$$ having $$m, n$$ as the least and the greatest elements respectively, is
Total number of subsets
= the number of selections of at least two elements including $$m, n$$ and natural numbers lying between $$m$$ and $$n$$
= total number of selections from $$n - m - 1$$ different things
$$ = {2^{n - m - 1}}.$$
20.
How many ways are there to arrange the letters in the word $$GARDEN$$ with vowels in alphabetical order
Total number of arrangements of letters in the word $$GARDEN = 6 ! = 720$$ there are two vowels $$A$$ and $$E,$$ in half of the arrangements $$A$$ preceeds $$E$$ and other half $$A$$ follows E.
So, vowels in alphabetical order in $$\frac{1}{2} \times 720 = 360$$
