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
A
216
B
240
C
600
D
3125
Answer :
216
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
A
36
B
90
C
120
D
None of these
Answer :
120
$$24 = 2 \cdot 3 \cdot 4,2 \cdot 2 \cdot 6,1 \cdot 6 \cdot 4,1 \cdot 3 \cdot 8,1 \cdot 2 \cdot 12,1 \cdot 1 \cdot 24$$
(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
A
105
B
315
C
630
D
450
Answer :
630
$$10 < _{7w}^{3m}3$$ 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
Answer :
$$\left( {p + m + n + 1} \right)\left( {n + 1} \right) - 1$$
$${3^p} \cdot {6^m} \cdot {21^n} = {2^m} \cdot {3^{p + m + n}} \cdot {7^n}$$
∴ 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
A
$$\frac{{6!8!10!}}{{4!6!}}$$
B
$$\frac{{8!8!10!}}{{4!6!}}$$
C
$$\frac{{8!8!6!}}{{6!4!}}$$
D
None of these
Answer :
$$\frac{{8!8!10!}}{{4!6!}}$$
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
A
270
B
220
C
282
D
None of these
Answer :
None of these
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
A
$${2^{n - m - 1}} - 1$$
B
$${2^{n - m - 1}}$$
C
$${2^{n - m}}$$
D
None of these
Answer :
$${2^{n - m - 1}}$$
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
A
480
B
240
C
360
D
120
Answer :
360
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 $$\frac{1}{2} \times 720 = 360$$
