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CBSE Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 4 - MCQExams.com
CBSE
Class 11 Engineering Maths
Permutations And Combinations
Quiz 4
If k is odd, then $$^k C_r$$ is maximum for r equal to
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0%
$$\dfrac{1}{2}(k-1)$$
0%
$$\dfrac{1}{2}(k+1)$$
0%
$$k-1$$
0%
$$k$$
Explanation
$$\because$$ Since $$k$$ is an odd number.
$$\therefore k=2n+1$$
Now, $${^kC_r}$$ becomes $${^{2n+1}C_r}$$
Now, condition for $${^{2n+1}C_r}$$ to be max is when $$r=n$$
but $$\therefore {^{n}C_r}={^{n}C_{2n+1-n}}={^{2n+1}C_{n+1}}$$
Or $$r=n$$ or $$r=n+1$$
but $$k=2n+1$$ , $$k=2n+1$$
or $$k=2r+1$$ , $$k=2(r-1)+1$$
$$\Rightarrow$$ $$r=\dfrac{k-1}{2}$$ , $$k=2r-1$$
$$\Rightarrow$$ , $$r=\dfrac{k+1}{2}$$
Hence, the answer is $$\dfrac{1}{2}({k-1}), \dfrac{1}{2}({k+1}).$$
What is the sum of all 5 digit numbers which can be formed with digits 0, 1, 2, 3, 4 without repetition
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0%
2599980
0%
2679980
0%
2544980
0%
2609980
Explanation
Total number of 5 digit numbers (including which begins with zero) $$=5!=120$$Number of 5 digit numbers which begin with zero $$=4!=24$$
Sum of all 5 digit numbers
$$=(0+1+2+3+4)\times 4!\times (11111)$$
$$=240\times 11111=2666640$$
Sum of all five digit numbers which begin with zero
$$=(1+2+3+4)\times 3!\times (1111)=66660$$
Hence the sum of the required numbers $$=sum $$ of all 5 digit numbers including those numbers
which begin with zero $$-sum $$ of all 5 digit numbers which begin with zero.
$$=2666640-66660=2599980$$
Given:
(i) $$^n p_r=\angle r ^ n c_r$$
(ii) $$^n c_r + ^n c_{r-1} = ^{n+1}c_r$$
If $$^n p_r = 720$$ and $$^n c_r = 120$$, then what is the value of 'r'
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0%
$$6$$
0%
$$9$$
0%
$$3$$
0%
$$\dfrac{n}{6}$$
Explanation
Given that $$_{ }^{ n }{ { P }_{ r } }=720$$ and $$_{ }^{ n }{ { C }_{ r } }=120$$,
We know that $$_{ }^{ n }{ { C }_{ r } }=\dfrac { n! }{ \left( n-r \right) !r! } $$ and $$_{ }^{ n }{ P_{ r } }=\dfrac { n! }{ \left( n-r \right) ! } $$,
$$\Longrightarrow \dfrac { _{ }^{ n }{ P_{ r } } }{ _{ }^{ n }{ { C }_{ r } } } =r!=6\\ \Longrightarrow r=3$$
How many four-letter computer passwords can be formed using only the symmetric letters (no repetition allowed)?
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$$7920$$
0%
$$330$$
0%
$$14640$$
0%
$$419430$$
Explanation
In English alphabets $$11$$ letters are there which are symmetric.
So, total ways $$=11\times10\times9\times8$$
$$=7920$$
Hence, the answer is $$7920.$$
In how many ways is it possible to choose a white square and a black square on a chessboards, so that the squares must not lie in the same row or column?
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$$56$$
0%
$$896$$
0%
$$60$$
0%
$$768$$
Explanation
In a chess board $$32$$ white squares are there and $$32$$ black.
If we choose $$32$$ white squares there will be $$8$$ of black square lying in same row or column of that white square.
So, for each white square we have $$24$$ choice.
$$\therefore$$ Overall $$=32\times24=768.$$
Hence, the answer is $$768.$$
Find the number of rectangles formed on a chessboard:
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1296
0%
1452
0%
1528
0%
1614
Explanation
In a chess board there are 8 columns and 8 rows.
Hence the number of rectangles
$$=1^{3}+2^{3}+3^{3}+....+8^{3}$$
$$=\left (\dfrac{8\times 9}{2} \right )^{2}=1296$$
If $$N$$ is the number of positive integral solutions of $$x_1\times x_2\times x_3\times x_4 = 770$$, then:
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0%
$$N$$ is divisible by $$4$$ district primes
0%
$$N$$ is a perfect square
0%
$$N$$ is a perfect fourth power
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$$N$$ is a perfect $$8^{th}$$ power
Explanation
(B, C, D) $$770=(2)(5)(7)(11).$$ We can assign each of $$2, 5, 7, 1$$ in $$4$$ ways. Thus, required number of solutions
$$4^{4}=256=16^{2}=2^{8}$$
Based on this information answer the questions
given
below.
A string of three English letters is formed as per the following rules:
(a) The first letter is any vowel.
(b) The second letter is $$m, n$$ or $$p$$.
(c) If the second letter is $$m$$ then the third letter is any vowel which is different from
the first letter.
(d) If the second letter is $$n$$ then the third letter is $$e$$ or $$u$$.
(e) If the second letter is $$p$$ then the third letter is the same as the first letter.
How many strings of letters can possibly be formed using the above rules?
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$$40$$
0%
$$45$$
0%
$$30$$
0%
$$35$$
Explanation
There are exclusive cases:
(i) when m is the second letter(ii) when n is the second letter(i) when p is the second letter
Case (i)
: First letter can be selected in 5 ways out of t vowels. Since second letter is fixed (i.e., m) therefore no. of
ways of selection of second letter is 1.
Third letter can be selected in 4 ways out of remaining 4 vowels (since the vowel which has been used at first place can
not be used at the place of third letter.
$$\therefore $$ Total no. of ways $$=5\times 1\times 4=20$$
Case (ii)
: First letter can be selected in 5 ways.Second letter can be selected in 1 way.Third letter can be selected in 1 way.Since third letter will be same as the first letter. So if the first letter is selected, then there is no need to select
the third letter.
$$\therefore $$ Total no. of ways $$=5\times 1\times 1=5$$
Hence , sum of all the possible no. of ways in which the string of letters can be formed
$$=20+10+5=35$$
On the occasion of Dipawali festival each student of a class sends greeting cards to one another. If the postmen deliver $$1640$$ greeting cards to the students of this class, then the number of students in the class is
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0%
39
0%
41
0%
51
0%
53
Explanation
let number of students is $$n$$, then number of cards $$=2\times ^nC_2= n(n-1) = 1640$$
Hence $$n = 41$$
State true or false.
The product of any $$r$$ consecutive natural numbers is always divisible by $$r!$$.
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0%
True
0%
False
Explanation
Let $$'r'$$ consecutive integers be $$x+1,x+2,...x+r$$
$$\displaystyle \therefore \left( x+1 \right) \left( x+2 \right) ...\left( x+r \right) =\frac { \left( x+r \right) \left( x+r-1 \right) ...\left( x+1 \right) ! }{ x! }$$
$$\displaystyle =\frac { \left( x+r \right) ! }{ \left( x \right) ! } .\frac { r! }{ r! } =^{ x+r }{ { C }_{ r } }.\left( r \right) !$$
Thus, $$\displaystyle \left( x+1 \right) \left( x+2 \right) ...\left( x+r \right) =^{ x+r }{ { C }_{ r } }.\left( r \right) !$$ which is clearly divisible by $$(r)!$$.
Hence, it is a true statement.
We number both the rows and the columns of an $$8$$ $$\times$$ $$8$$ chess-board with the numbers $$1$$ to $$8$$. A number of grains are placed onto each square, in such a way that the number of grains on a certain square equals the product of its row and column numbers. How many grains are there on the entire chessboard?
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$$1296$$
0%
$$1096$$
0%
$$2490$$
0%
$$1156$$
Explanation
In the first column, we have, successively,
$$1 \times 1, 1 \times 2, 1 \times 3, ..... , 1 \times 8$$ grains.
So, in the first column : $$ 1 \times (1+2+3+4+5+6+7+8)$$
In the second column : $$ 2 \times (1+2+3+4+5+6+7+8)$$
In the third column : $$ 3 \times (1+2+3+4+5+6+7+8).$$
Finally, in the eighth column : $$ 8 \times (1 + 2 + 3+ 4+5+6+7+8)$$
Since $$1+2+3+4+5+6+7+8 = (1/2)\times 8 \times(1+8) = 36$$, there are $$36^2 =1296$$ grains on the board.
In how many ways is it possible to choose a white square and a black square on a chess board so that the squares must not lie in the same row or column -
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0%
$$56$$
0%
$$896$$
0%
$$60$$
0%
$$768$$
Explanation
We know a chess board has $$32$$ white and $$32$$ black squares, we need to select $$1$$ white square out of these $$32$$ white squares that is done in $$_{ }^{ 32 }{ C }_{ 1 }^{ }{ }$$ which is $$32$$
No, we have selected any one white square, there are eight black squares lying in the same column or row so out of $$32-8=24$$ black squares
We need to select one black square that is $$_{ }^{ 24 }{ C }_{ 1 }^{ }{ }$$
So, if both squares selected then our work is done
Hence it is in $$32\times 24=768$$ ways.
A hall has 12 gates. In how many ways can a man enter the hall through one gate and come out through a different gate?
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$$144$$
0%
$$132$$
0%
$$121$$
0%
$$156$$
Explanation
The person can enter the hall in 12 gates and exit from any of the other 11 gates.
Hence the total number of ways $$ = 12 \times 11 =132 $$
$$\displaystyle \frac{^{\alpha}C_{2}}{^{\alpha + 1}C_{4}}$$ =
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0%
$$\displaystyle\frac{1}{3}$$
0%
$$\displaystyle\frac{12}{(\alpha+1)(\alpha-2)}$$
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$$\displaystyle\frac{4}{(\alpha+1)(\alpha-3)}$$
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$$\displaystyle\frac{12}{(\alpha+1)(\alpha-2)(\alpha-3)}$$
Explanation
$$=\dfrac{^{\alpha}C_2}{^{\alpha + 1}C_4}$$
$$=\dfrac{\alpha !4!(\alpha -3)!}{(\alpha -2)!2!(\alpha +1)!}$$
$$=\dfrac{\alpha !4.3.2!(\alpha -3)!}{(\alpha -2)(\alpha -3)!2!(\alpha +1)\alpha !}$$
$$=\dfrac{12}{(\alpha -2)(\alpha +1)}$$
Hence, option 'B' is correct.
There are three stations $$A,\space B$$ and $$C$$, five routes for going from station $$A$$ to station $$B$$ and four routes for going from station $$B$$ to station $$C$$.
Find the number of different ways through which a person can go from station $$A$$ to $$C$$ via $$B$$
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0%
10
0%
15
0%
20
0%
25
Explanation
Since there are five routes for going from station A to station B
So number of ways of going from A to B is $$5$$
And there are four routes for going from B to C, so the number of ways of going from B to C is $$4$$
Thus using multiplication rule, number of ways of going from A to C via B is, $$5\times 4=20$$
Given six line segments of length $$2, 3, 4, 5, 6, 7$$ units. Then the number of triangles that can be formed by joining these lines is
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0%
$$^6C_3 - 7$$
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$$^6C_3 - 1$$
0%
$$^6C_3$$
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$$^6C_3 - 2$$
Explanation
Given lengths of line segments are $$2,3,4,5,6,7$$,
The number ways of selecting $$3$$ lengths from them is $$_{ }^{ 6 }{ { C }_{ 3 } }$$,
Consider the cases in which a triangle is not formed,
i.e., when any one of these two conditions is not satisfied,
$$1)$$sum of any two lengths is greater than the third length
$$2)$$difference between any two lengths is smaller than the third one,
Now,
number of such cases are
$$2,3,5$$
$$2,3,6$$
$$2,3,7$$
$$2,4,6$$
$$2,4,7$$
$$2,5,7$$
$$3,4,7$$
$$\therefore$$ the total number of ways are $$_{ }^{ 6 }{ { C }_{ 3 } }-7$$
The number of different ways in which five "alike dashes" and eight "alike dots" can be arranged using only seven of these "dashes" and "dots" is
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$$350$$
0%
$$120$$
0%
$$1287$$
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None of these
Explanation
Dashes
Dots
Arrangements
$$5$$
$$2$$
$$^7C_2$$
$$4$$
$$3$$
$$^7C_3$$
$$3$$
$$4$$
$$^7C_4$$
$$2$$
$$5$$
$$^7C_5$$
$$1$$
$$6$$
$$^7C_6$$
$$0$$
$$7$$
$$^7C_7$$
The total number of ways is $$^7C_2+^7C_3+^7C_4+^7C_5+^7C_6=2^7-8=120$$.
Number of ways 6 rings can be worn on four fingers of one hand?
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$$4095$$
0%
$$4096$$
0%
$$4097$$
0%
$$4098$$
Explanation
These are rings so, repetition is allowed
Number of ways for first ring are $$4$$, for two rings $$4\times 4$$...
So for $$6$$ rings there are $$4^{6} ways = 4096$$ ways
Hence, option 'B' is correct.
The value of $$\displaystyle ^{ 47 }{ { C }_{ 4 } }+\sum _{ r=1 }^{ 5 }{ ^{ 52-r }{ { C }_{ 3 } } } $$ is equal to
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$$^{ 47 }{ { C }_{ 6 } }$$
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$$^{ 52 }{ { C }_{ 5 } }$$
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$$^{ 52 }{ { C }_{ 4 } }$$
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none of these
Explanation
We need value of $$={ _{ }^{ 47 }{ C } }_{ 4 }^{ }+\sum _{ r=1 }^{ 5 }{ { _{ }^{ 52-r }{ C } }_{ 3 }^{ } } $$
$$={ _{ }^{ 47 }{ C } }_{ 4 }^{ }+{ _{ }^{ 51 }{ C } }_{ 3 }^{ }+{ _{ }^{ 50 }{ C } }_{ 3 }^{ }+{ _{ }^{ 49 }{ C } }_{ 3 }^{ }+{ _{ }^{ 48 }{ C } }_{ 3 }^{ }+{ _{ }^{ 47 }{ C } }_{ 3 }^{ }$$
$$={ _{ }^{ 51 }{ C } }_{ 3 }^{ }+{ _{ }^{ 50 }{ C } }_{ 3 }^{ }+{ _{ }^{ 49 }{ C } }_{ 3 }^{ }+{ _{ }^{ 48 }{ C } }_{ 3 }^{ }+{ _{ }^{ 47 }{ C } }_{ 3 }^{ }+{ _{ }^{ 47 }{ C } }_{ 4 }={ _{ }^{ 51 }{ C } }_{ 3 }^{ }+{ _{ }^{ 50 }{ C } }_{ 3 }^{ }+{ _{ }^{ 49 }{ C } }_{ 3 }^{ }+{ _{ }^{ 48 }{ C } }_{ 3 }^{ }+{ _{ }^{ 48 }{ C } }_{ 4 }^{ }$$
$$={ _{ }^{ 51 }{ C } }_{ 3 }^{ }+{ _{ }^{ 50 }{ C} }_{ 3 }^{ }+{ _{ }^{ 49 }{ C } }_{ 3 }^{ }+{ _{ }^{ 49 }{ C } }_{ 4 }^{ }={ _{ }^{ 51 }{ C } }_{ 3 }^{ }+{ _{ }^{ 50 }{ C } }_{ 3 }^{ }+{ _{ }^{ 50 }{ C } }_{ 4 }^{ }$$
$$={ _{ }^{ 51 }{ C } }_{ 3 }^{ }+{ _{ }^{ 51 }{ C } }_{ 4 }^{ }={ _{ }^{ 52 }{ C } }_{ 4 }^{ }$$
If $$m$$ parallel lines in a plane are intersected by a family of $$n$$ parallel lines, the number of parallelograms than can be formed is
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0%
$$\dfrac {1}{4} mn(m - 1)(n - 1)$$
0%
$$\dfrac {1}{2} mn (m - 1)(n - 1)$$
0%
$$\dfrac {1}{4} m^{2}n^{2}$$
0%
None of these
Explanation
The number of selection of two parallel from $$m$$ lines is $$^{m}C_{2}$$.
The number of selection of two parallel lines from $$n$$ lines is $$^{n}C_{2}$$.
Hence, the number of parallelograms lines is
$$^{m}C_{2} \times ^{n}C_{2} = \dfrac {1}{4} mn (m - 1)(n - 1)$$.
The total number of permutation of $$n$$ different things taken not more than $$r$$ at a time, where each thing may be repeated any number of times, is
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0%
$$\displaystyle \frac { n\left( { n }^{ n }-1 \right) }{ n-1 } $$
0%
$$\displaystyle \frac { \left( { n }^{ r }-1 \right) }{ n-1 } $$
0%
$$\displaystyle \frac { n\left( { n }^{ r }-1 \right) }{ n-1 } $$
0%
$$\displaystyle \frac { \left( { n }^{ n }-1 \right) }{ n-1 } $$
Explanation
One place can be filled in $$n$$ ways by $$n$$ things, the first and second places can be filled in $$=n\times n={ n }^{ 2 }$$ ways.
The first, second and third places can be filled in $$=n\times n\times n={ n }^{ 3 }$$
Similarly, the first, second,....rth places can be filled in $$=n\times n\times n\times ......\times n\left( r\ times \right) ={ n }^{ r }$$
So required total number $$=n+{ n }^{ 2 }+{ n }^{ 3 }+.....+{ n }^{ r }=\displaystyle\frac { n\left( { n }^{ r }-1 \right) }{ n-1 } $$ (Sum of a GP)
The maximum number of points of intersection of five lines and four circles is
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0%
$$60$$
0%
$$72$$
0%
$$62$$
0%
None of these
Explanation
Two circles intersect at two distinct points. Two straight lines intersect at one point.
One circle and one straight line intersect at two distinct points.
Then the total numbers of points of intersections are as follows:
Number of ways of selection
Points of intersection
Two straight lines: $$^5C_2$$
$$^5C_2\times 1=10$$
Two circle: $$^4C_2$$
$$^4C_2\times 2=12$$
One line and one circle: $$^5C_1\times ^4C_1$$
$$^5C_1\times ^4C_1\times 2=40$$
Total
$$= 62$$
If $$\displaystyle ^nC_{r-1}=56$$, $$\displaystyle ^nC_r=28$$ and $$\displaystyle ^nC_{r+1}=8$$ then r is equal to
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0%
8
0%
6
0%
5
0%
None of these
Explanation
From the given information, $$\dfrac{\:^{n}C_{r-1}}{\:^{n}C_{r}}=2$$
$$\Rightarrow \dfrac{n!r!(n-r)!}{(r-1)!(n-(r-1))!.n!}=2$$
$$\Rightarrow \dfrac{r}{n+1-r}=2$$
$$\Rightarrow r=2n+2-2r$$
$$\Rightarrow 3r=2n+2$$...(i)
Also, $$\dfrac{\:^{n}C_{r}}{\:^{n}C_{r+1}}=\dfrac{7}{2}$$
$$\Rightarrow \dfrac{n!(n-(r+1))!.(r+1)!}{n!.r!.(n-r)!}=\dfrac{7}{2}$$
$$\Rightarrow \dfrac{r+1}{n-r}=\dfrac{7}{2}$$
$$\Rightarrow 2r+2=7n-7r$$
$$\Rightarrow 9r=7n-2$$ ...(ii)
$$\Rightarrow n=8$$
Hence,
$$\Rightarrow 9r=56-2$$
$$\Rightarrow 9r=54$$
$$\Rightarrow r=6$$
If $$^nC_{r-1} = 36, \space ^nC_r = 84$$ and $$^nC_{r+1} = 126,$$ then find $$r$$.
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0%
2
0%
3
0%
4
0%
5
Explanation
Here
$$\displaystyle\frac{^nC_r}{^nC_{r-1}} = \displaystyle\frac{84}{36}$$
$$\Rightarrow
\displaystyle\frac{n-r+1}{r} = \displaystyle\frac{7}{3} \quad\quad
\left(\because \quad \displaystyle\frac{^nC_r}{^nC_{r-1}}
= \displaystyle\frac{n-r+1}{r} \right)$$
$$\Rightarrow 3n-3r+3 = 7r$$
$$\Rightarrow 10r - 3n = 3 \quad\quad (i)$$
And
$$\displaystyle\frac{^nC_{r+1}}{^nC_r} =
\displaystyle\frac{n- (r+1) +1}{(r+1)}
= \displaystyle\frac{126}{84} \quad\quad \left(\because
\quad \displaystyle\frac{^nC_r}{^nC_{r-1}}
= \displaystyle\frac{n-r+1}{r} \right)$$
$$\Rightarrow \displaystyle\frac{n-r}{r+1} = \displaystyle\frac{3}{2}$$
$$\Rightarrow 2n - 2r = 3r + 3$$
$$\Rightarrow 5r - 2n = -3$$
$$\Rightarrow 10r-4n = -6 \quad\quad ...(ii)$$
Subtracting $$(ii)$$ from $$(i)$$, we get $$n = 9$$
from $$(i),\quad 10r - 27 = 3$$
$$\Rightarrow 10r = 30$$
$$\therefore \quad\quad r = 3$$.
The number of numbers of 9 different nonzero digits such that all the digits in the first four places are less than the digit in the middle and all the digits in the last four places are greater than that in the middle is
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0%
$$\displaystyle 2(4!)$$
0%
$$\displaystyle (4!)^2$$
0%
$$8!$$
0%
none of these
Explanation
$$\textbf{Step 1: Given conditions and finding the middle digit.}$$
$$\text{Given:}$$
$$\text{Numbers of 9 different nonzero digits with following conditions}$$
$$\text{Condition 1:Digits in first four places should be less than the middle digit.}$$
$$\text{Condition 2:Digits in last four places should be greater than the middle digit.}$$
$$\text{According to the conditions the middle digit will be 5; first four digits will be 1,2,3,4 and the last four}$$
$$\text{digits will be 6,7,8,9.}$$
$$\textbf{Step 2: Permutations of first 4 digits and last 4 digits.}$$
$$\text{The first four digits can be permutated with 1,2,3,4}$$
$$\text{Hence the number of Permutations will be}$$ $${4!}$$
$$\text{The last four digits can be permutated with 6,7,8,9}$$
$$\text{Hence the number of Permutations will be}$$ $${4!}$$
$$\textbf{Step 3: Final calculation.}$$
$$\text{Therefore the number of numbers of 9 different nonzero digits is}$$
$$\Rightarrow \text{4!.4!}$$
$$\Rightarrow {(4!)^2}$$
$$\textbf{Hence option B is correct.}$$
In a packet there are m different books, n different pens and p different pencils. The number of selections of at least one article of each type from the packet is
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0%
$$\displaystyle 2^{m+n+p}-1$$
0%
$$\displaystyle \left ( m+1 \right )\left ( n+1 \right )\left ( p+1 \right )-1$$
0%
$$\displaystyle 2^{m+n+p}$$
0%
$$(2^{m}-1)(2^{n}-1)(2^{p}-1)$$
Explanation
Since each of $$(m+n+p)$$ items has 2 choices of being selected or not selected; hence total no. of ways of selecting items $$=2\times 2\times 2\times 2\times .................(m+n+p)times={ 2 }^{ m+n+p }$$
but one case in this will be when none of the items is selected.
Therefore, no. of ways of selecting at least one item of each type
$$(2^{m}-1)(2^{n}-1)(2^{p}-1)$$
Hence, option 'D' is correct.
The value of $$\displaystyle \sum_{r= 0}^{n}$$ $$\displaystyle ^{n+r}C_{r}$$ is equal to
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0%
$$\displaystyle ^{2n+1}C_{n}$$
0%
$$\displaystyle ^{2n}C_{n-1}$$
0%
$$\displaystyle ^{2n}C_{n+1}$$
0%
$$\displaystyle ^{2n+1}C_{n-1}$$
Explanation
The given series can be written as
$$^{ n }{ C }_{ 0 }+^{ n+1 }{ C }_{ 1 }+^{ n+2 }{ C }_{ 2 }+.....+^{ 2n }{ C }_{ n }\\ =\left\{ ^{ n+1 }{ C }_{ 0 }+^{ n+1 }{ C }_{ 1 } \right\} +^{ n+2 }{ C }_{ 2 }+.....+^{ 2n }{ C }_{ n }\\ =\left\{ ^{ n+2 }{ C }_{ 1 }+^{ n+2 }{ C }_{ 2 } \right\} +.....+^{ 2n }{ C }_{ n }\\ =\left\{ ^{ n+3 }{ C }_{ 2 }+^{ n+3 }{ C }_{ 3 } \right\} +^{ n+4 }{ C }_{ 4 }+....+^{ 2n }{ C }_{ n }\\ =^{ n+4 }{ C }_{ 3 }+^{ n+4 }{ C }_{ 4 }+....+^{ 2n }{ C }_{ n }$$
Proceeding the same way finally we get the sum as
$$^{ 2n+1 }{ C }_{ n }$$
The number of 5-digit even numbers that can be made with the digits 0,1,2 and 3 is
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0%
384
0%
192
0%
768
0%
none of these
Explanation
Here even number must end with $$0$$ or $$2$$
Since $$5$$ digit number is to be formed with $${0,1,2,3}$$ hence it is obvious that repetition is allowed.
$$\left< { a }|{ b }|{ c }|{ d }|{ e } \right> $$
here "$$a$$" has $$3$$ options $${1,2,3}$$ ,$$b,c,d$$ has $$4$$ options
$${0,1,2,3}$$ and "$$e$$" has $$2$$ options $${0} $$ and $${2}$$
Number of $$5$$ digit numbers = $$3\times 4\times 4\times 4\times 2=384$$
Hence, option 'A' is correct.
If $$\displaystyle ^nC_4,^nC_5 $$ and $$ ^nC_6$$ are in AP then n is
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0%
8
0%
9
0%
14
0%
7
Explanation
If $$\displaystyle ^nC_4,^nC_5 $$ and $$ ^nC_6$$ are in A.P.
$$\Rightarrow 2.^nC_5 = ^nC_4+^nC_6$$
$$\Rightarrow \cfrac{2}{5}=\cfrac{1}{(n-4)}+\cfrac{n-5}{6}$$
$$\Rightarrow n^2-21n+98=0\Rightarrow n = 7,14$$
From a group of persons the number of ways of selecting 5 persons is equal to that of 8 persons. The number of persons in the group is
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0%
13
0%
40
0%
18
0%
21
Explanation
we know, number of ways of selecting r persons from n persons =$$^{n}\textrm{C}_{r}$$
i.e. $$^{n}\textrm{C}_{5}$$= $$^{n}\textrm{C}_{8}$$
but, $$^{n}\textrm{C}_{r}$$=$$^{n}\textrm{C}_{n-r}$$
=> $$n=r+(n-r)$$
=> $$n=5+8$$
=> $$n=13$$
$$\displaystyle ^{n}C_{r+1}+^{n}C_{r-1}+2\times ^{n}C_{r}$$ is equal to
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0%
$$\displaystyle ^{n+2}C_{r+1}$$
0%
$$\displaystyle ^{n+1}C_{r}$$
0%
$$\displaystyle ^{n+1}C_{r+1}$$
0%
$$\displaystyle ^{n+2}C_{r}$$
Explanation
$$\:^{n}C_{r+1}+\:^{n}C_{r-1}+\:^{n}C_{r}+\:^{n}C_{r}$$
$$=(\:^{n}C_{r+1}+\:^{n}C_{r})+(\:^{n}C_{r}+\:^{n}C_{r-1})$$
$$=\:^{n+1}C_{r+1}+\:^{n+1}C_{r}$$
$$=\:^{n+2}C_{r+1}$$
This comes from the property that
$$\:^{n}C_{r-1}+\:^{n}C_{r}=\:^{n+1}C_{r}$$
The number of different 6-digit numbers that can be formed using the three digits 0,1 and 2 is
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0%
$$\displaystyle 3^6$$
0%
$$\displaystyle 2$$ x $$3^5$$
0%
$$\displaystyle 3^5$$
0%
none of these
Explanation
in the six places of the digits, the first place can be filled only by $$0$$1 or $$2$$ (if $$0$$ comes in first place, that makes it $$5$$ digit number) and the remaining places can be filled by any one of the $$3$$ digits.
So, first place has only $$2$$ choices and the remaining $$5$$ places have $$3$$ choices each.
Hence,the number of different $$6$$-digit numbers that can be formed using the three digits $$0,1$$ and $$2$$ = $$2 \times 3^{5}$$
The product of r consecutive integers is divisible by
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0%
r
0%
$$\displaystyle \sum_{k=1}^{r-1}k$$
0%
r!
0%
none of these
Explanation
Product of $$r$$ consecutive integers is always divisible by $$=r!$$
Hence it is divisible by $$r$$ (option A), $$r!$$ (option C) and $$r(r-1)/2$$ (option B)
The total number of $$9-$$digit numbers of different digits is
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0%
$$10 (9!)$$
0%
$$8 (9!)$$
0%
$$9 (9!)$$
0%
none of these
Explanation
All number are formed by using the digits $$0,1,2,3,4,5,6,7,8,9$$
The number of arrangements each consisting of $$9$$ digits $$=^{ 10 }{ P }_{ 9 }=10!$$
These arrangements includes those which start from $$0$$,
clearly all such numbers will be $$8$$ digit number.
the number of such arrangement $$=^{ 9 }{ P }_{ 8 }=9!$$
Therefor number of all $$9$$ digit number $$=10!-9!=10\left( 9! \right) -9!=9\times 9!$$
For $$\displaystyle 2\leq r\leq n,\binom{n}{r}+2\binom{n}{r-1}+\binom{n}{r-2}$$ is equal to
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0%
$$\displaystyle \binom{n+1}{r-1}$$
0%
$$\displaystyle 2\binom{n+1}{r+1}$$
0%
$$\displaystyle 2\binom{n+2}{r}$$
0%
$$\displaystyle \binom{n+2}{r}$$
Explanation
$$\:^{n}C_{r}+\:^{n}C_{r-1}+\:^{n}C_{r-1}+\:^{n}C_{r-2}$$
$$=(\:^{n}C_{r}+\:^{n}C_{r-1})+(\:^{n}C_{r-1}+\:^{n}C_{r-2})$$
$$=\:^{n+1}C_{r}+\:^{n+1}C_{r-1}$$
$$=\:^{n+2}C_{r}$$
This comes from the property that
$$\:^{n}C_{r-1}+\:^{n}C_{r}=\:^{n+1}C_{r}$$
The numbers of three digit number can be formed by using the digits 0, 1, 2, 3, 7, 9 if any digit can be used in any number of time, is
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0%
180
0%
216
0%
120
0%
36
Explanation
Consider the 3 digit number as
- - -
The total number of digits available are $$0,1,2,3,7,9$$.
Hence the number of ways in which we can fill the first place will be
$$6-1$$
$$=5$$ ...(excluding 0).
Similarly the second place can be filled in $$6$$ ways.
The third place can be filled in $$ 6$$ ways.
Hence in total there will be
$$5\times 6\times 6$$ numbers
$$=36\times 5$$
$$=180$$.
If $$n<p<2n$$ and $$p$$ is prime and $$N=^{ 2n }{ { C }_{ n } }$$, then
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0%
$$\displaystyle \frac { p }{ N } $$
0%
$$p$$ does not divide $$N$$
0%
$$\displaystyle \frac { p^2 }{ N } $$
0%
$$p^2$$ does not divide $$N$$
Explanation
A,D
$$N=\displaystyle\frac { 2n! }{ n!n! } =\frac { \left( n+1 \right) \left( n+2 \right) ....\left( 2n \right) }{ n! } \Rightarrow n!N=\left( n+1 \right) \left( n+2 \right) ....\left( 2n \right) $$
$$\displaystyle\frac { p }{ \left( n+1 \right) \left( n+2 \right) ....\left( 2n \right) } $$
as $$n\le p\le 2n$$
Thus $$\dfrac { p }{ n!N } $$ or
$$\dfrac { p }{ N } $$ as n! does not divide p.
If possible, let $$\displaystyle\dfrac { { p }^{ 2 } }{ N } $$
$$\Rightarrow \displaystyle\dfrac { { p }^{ 2 } }{ n!N } =\dfrac { { p }^{ 2 } }{ \left( n+1 \right) \left( n+2 \right) ....\left( 2n \right) } $$
or $$\dfrac { p }{ \left( n+1 \right) \left( n+2 \right) ....\left( 2n \right) } $$
This is not possible as $${p }^{ 2 }$$ does not divide N.
so A D hold.
The number of ways in which one or more letters be selected from the letters $$AAAABBCCCDEF$$ is
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0%
$$476$$
0%
$$487$$
0%
$$435$$
0%
$$479$$
Explanation
The number of ways of selecting
$$A=4+1=5, B=2+1=3, C=3+1=4,D=1+1=2, E=1+1=2,F=1+1=2$$
So total ways of selecting is $$5\times 4\times 3\times 2\times 2\times 2=480$$
Above count also includes the one count when no letter is selected. we need cases when atleast one letter is selected so number of ways$$=480-1=479$$
The letters of the word COCHIN are permuted and all the permutations are arranged in an alphabetical order as in an English dictionary. The number of words that appear before the word COCHIN is
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0%
360
0%
192
0%
96
0%
48
Explanation
Letters appearing in the word COCHIN are
C,C,H,I,N,O
the words include the words of the form CC---- , CH---- , CI----, CN---- and the first words starting with CO---- is COCHIN
thus,the number of words before COCHIN is $$(4)(4!)=96$$
Hence Option $$C$$ is correct
The number of even numbers with three digits such that if 3 is one of the digit then 5 is the next digit are
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0%
959
0%
285
0%
365
0%
512
Explanation
Number of 3 digit even numbers with no $$3=8\times 9\times 5=360$$
$$8$$ because 1st place can't be 0 and 3 is not included.
No of 3 digit even number with 3 and hence the constraints$$=1\times 1\times 5=5$$
note that 2nd and 3rd position can't be 3 since in that case the number can't be even and fulfilling constraint at the same time .
Therefore, the number of such numbers $$=(360+5)=365$$
Hence, option 'C' is correct.
The different six digit numbers whose 3 digits are even and 3 digits are odd is
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0%
281250
0%
281200
0%
156250
0%
none of these
Explanation
If first digit is even then select other places from remaining five places for two more even digits.
This can be $$\displaystyle 4\times ^{5}C_{2}\times 5\times 5.$$
Now the remaining three places are to be fit up by odd digit which can be done in $$\displaystyle 5^{3}$$ ways.
In this way number of six digit number in which three digit are even and $$3$$ are odd in which three digit are even and $$3$$ are odd in which the first digit is even
$$=\displaystyle 4\times ^{5}C_{2}\times 5\times 5\times 5^{3}=8\times 5^{6}$$ (0 can't occurs at first place)
Similarly number of six digit numbers which begin with an odd number/digit and which have $$3$$ even, $$3$$ odd digits
$$\displaystyle =5\times 5C_{2}\times 5\times 5\times 5^{3}=5\times10\times 5^{5}=10.5^{6}$$
$$\displaystyle \therefore $$ Total six digit number in which $$3$$ odd $$3$$ even are
$$\displaystyle =8\times 5^{6}+10\times 5^{6}=18\times 5^{6}=281250$$
A five-digit number divisible by $$3$$ is to be formed using the digits $$0, 1, 2, 3, 4$$ and $$5$$, without repetition. The total number of ways this can be done is
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0%
$$216$$
0%
$$240$$
0%
$$600$$
0%
$$3125$$
Explanation
We know that a number is divisible by $$3$$ if the sum of its digits is divisible 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:
Total number of five-digit numbers formed using the digits $$1, 2, 3, 4, 5$$ is $$5! = 120$$.
Case II:
Taking $$0, 1, 2, 4, 5$$, total number is $$4\times 4! = 96$$.
From case I and case II, total number divisible by $$3$$ is $$120 + 96 = 216$$.
If $$m=^{n}\textrm{C}_{2}$$, then $$^{m}\textrm{C}_{2}$$ equals
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0%
$$^{n+1}\textrm{C}_{4}$$
0%
$$3\times ^{n+1}\textrm{C}_{4}$$
0%
$$^{n}\textrm{C}_{4}$$
0%
$$^{n+1}\textrm{C}_{3}$$
Explanation
$$\displaystyle m=^{n}C_{2}=\frac{n\left ( n-1 \right )}{2}...\left ( \ast \right )$$
$$\displaystyle
\therefore ^{m}C_{2}=\frac{m\left ( m-1 \right )}{2}=\frac{\frac{n\left
( n-1 \right )}{2}\left [ \frac{n\left ( n-1 \right )}{2}-1 \right
]}{2}$$ using (*)
$$\displaystyle =\frac{n\left ( n-1 \right )\left (
n^{2}-n-2 \right )}{8}=\frac{n\left ( n-1 \right )\left ( n-2 \right
)\left ( n+1 \right )}{8}$$
$$=\dfrac{3\left ( n+1 \right )\left ( n \right
)\left ( n-1 \right )\left ( n-2 \right )}{4!}$$
$$ =3. ^{n+1}C_{4}$$
The domain and range of the function $$\displaystyle f\left ( x \right ) = \sqrt{^{x^{2}+4x}C_{2x^{2}+3}}$$ are
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0%
domain: $$1,2,3 $$, range: $$1,2\sqrt { 3 } $$
0%
domain: $$1,2\sqrt {3} $$ range: $$1,2,3$$
0%
domain: $$1,3$$ range: $$1$$
0%
none of these
Explanation
$$\displaystyle f\left ( x \right ) = \sqrt{^{x^{2}+4x}C_{2x^{2}+3}}$$
For this function to exist $$x^2+4x \geq 0 \Rightarrow x \in R - [-4,0]$$
and $$x^2+4x\geq 2x^2+3\Rightarrow x^2-4x+3 \leq 0 \Rightarrow (x-3)(x-1) \leq 0 \Rightarrow x = 1,2,3$$
Hence domain is $$1,2,3$$ and range is $$\sqrt{^5C_5}, \sqrt{^{13}C_{11}}, \sqrt{^{21}C_{21}} = 1, 2\sqrt{3}$$
If $$^{ n }{ { C }_{ r-1 } }=\left( { k }^{ 2 }-8 \right) \left( ^{ n+1 }{ { C }_{ r } } \right) $$, then $$k$$ belongs to
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0%
$$[-3,-2\sqrt{2}]$$
0%
$$[-3,-2\sqrt{2})$$
0%
$$2\sqrt{2}.3]$$
0%
$$(2\sqrt{2},3]$$
Explanation
Given,
$$^{ n }{ { C }_{ r-1 } }=\left( { k }^{ 2 }-8 \right) \left( ^{ n+1 }{ { C }_{ r } } \right) $$
We have $$r-1\ge 0,r\le n+1$$
Therefore $$1\le r\le n+1\Rightarrow \displaystyle\frac { 1 }{ n+1 } \le \frac { r }{ n+1 } \le 1$$
Also, $${ k }^{ 2 }-8=\displaystyle\frac { n! }{ \left( n-r \right) !\left( n-r+1 \right) ! } =\frac { r }{ n+1 } $$
$$\Rightarrow \displaystyle\frac { 1 }{ n+1 } \ge { k }^{ 2 }-8\le 1$$
$$\Rightarrow 8<\displaystyle\frac { 1 }{ n+1 } +8\le { k }^{ 2 }\le 9$$
$$\Rightarrow 8<{ k }^{ 2 }\le 9$$
so,
$$k\in [-3, -2 \sqrt 2] \ and \ k \in \left[ 2\sqrt { 2 } ,3 \right] $$
The mean of the values $$0, 1, 2, \cdots n$$, having corresponding weights $$^nC_0, ^nC_1, \cdots ^nC_n,$$ respectively is
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0%
$$\cfrac{n+1}{2}$$
0%
$$\cfrac{2^{n-1}+1}{n+1}$$
0%
$$\cfrac{2^n}{n}$$
0%
$$\cfrac{n}{2}$$
Explanation
Required A.M $$=\cfrac{\sum x_if_i}{\sum f_i} = \cfrac{0.^nC_0+^nC_1+2.
^nC_2+3. ^nC_3+..............+n. ^nC_n}{^nC_0+^nC_1+ ^nC_2+
^nC_3+.......+ ^nC_n} = \mu $$ (say)
Now consider $$(1+x)^n =^nC_0+^nC_1 x+ ^nC_2 x^2+.........+^nC_n x^n ....(1)$$
Differentiating both side (1) w.r.t $$x$$
$$n(1+x)^{n-1} =^nC_1 +2. ^nC_2 x+.........+n. ^nC_n x^{n-1} ....(2)$$
Now putting $$x=1$$ in (1) and (2) we get,
$$^nC_0+^nC_1 + ^nC_2 +.........+^nC_n =2^n$$
and $$ ^nC_1 +2. ^nC_2 +.........+n. ^nC_n = n. 2^{n-1}$$
$$\therefore \mu = \cfrac{n. 2^{n-1}}{2^n}=\cfrac{n}{2}$$
An $$n-$$ digit number is a positive number with exactly $$n$$ digits.
Nine hundred distinct n−
digit
numbers are to be formed using only the three digits 2,5 and 7.
The smallest value of $$n$$ for which this is possible is
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0%
$$5$$
0%
$$6$$
0%
$$7$$
0%
$$8$$
Explanation
Using $$2,5$$ and $$7$$ with repetition each place of $$n$$ digit number can be chosen in $$3$$ ways.
Hence, total number of $$n-$$digit numbers $$=3\times 3\times 3...n$$ times $$={3}^{n}.$$
According to given condition $${ 3 }^{ n }\ge 900\Rightarrow { 3 }^{ n-2 }\ge 100$$
$$\therefore n-2\ge 5\Rightarrow n\ge 7$$
If $$^{20}C_{r\, +\, 2}\, =\, ^{20}C_{2r\, -\, 3}\,$$, find $$\, ^{12}C_r$$.
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0%
$$792$$
0%
$$795$$
0%
$$790$$
0%
None of these
Explanation
$$^{20}C_{r+2} = ^{20}C_{2r-3}$$
$$\Rightarrow r+2 = 2r-3$$, or$$ r+2=20-2r+3$$ [ Using $$^nC_x=^nC_y=> x=y\ or x=n-y$$ ]
$$\Rightarrow r = 5 \ or \ r=7 \therefore ^{12}C_5 = ^{12}C_7=792$$
A set contains $$(2n + 1)$$ elements. The number of subsets of the set which contain at most n elements is
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0%
$$2^{n}$$
0%
$$2^{n\, +\, 1}$$
0%
$$2^{n\, -\, 1}$$
0%
$$2^{2n}$$
Explanation
Let $$N$$ = the number of subsets of the set which contain at most n elements
$$=^{ 2n+1 }{ { C }_{ 0 } }+^{ 2n+1 }{ { C }_{ 1 } }+^{ 2n+1 }{ { C }_{ 2 } }+...+^{ 2n+1 }{ { C }_{ n } }$$
$$ 2N=2\left( ^{ 2n+1 }{ { C }_{ 0 }+^{ 2n+1 }{ { C }_{ 1 }+^{ 2n+1 }{ C_{ 2 }+...+^{ 2n+1 }{ { C }_{ n } } } } } \right) $$
$$ =\left( ^{ 2n+1 }{ { C }_{ 0 }+^{ 2n+1 }{ { C }_{ 2n+1 } } } \right) +\left( ^{ 2n+1 }{ { C }_{ 1 }+ }^{ 2n+1 }{ { C }_{ 2n } } \right) +...+\left( ^{ 2n+`1 }{ { C }_{ n }+^{ 2n+1 }{ { C }_{ n+1 } } } \right) $$........................
$$ \left( \because ^{ n }{ { C }_{ r }= }^{ n }{ C }_{ n-r } \right) $$
$$ =^{ 2n+1 }{ C }_{ 0 }+^{ 2n+1 }{ C }_{ 1 }+^{ 2n+1 }{ { C }_{ 2 } }+...+^{ 2n+1 }{ { C }_{ 2n+1 } }$$
$$ { =2 }^{ 2n+1 }$$
$$ \Rightarrow 2N={ 2 }^{ 2n+1 }$$
$$ \Rightarrow N={ 2 }^{ 2n }$$
The number of such numbers which are divisible by two and five (all digits are not different) is
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0%
125
0%
76
0%
65
0%
100
Explanation
The numbers which are divisible by 2 and 5 are the multiples of 10.
Hence the ending digit has to be 0.
However the starting digit cannot be 0.
Now
_ _ _ 0.
Hence number of ways of filling the first place will be 4, the second place will be 5 and 3rd place will be 5, since repetition of digits are allowed.
Hence total number of the numbers which are divisible by 2 and 5 are
$$4\times 5\times 5$$
$$=100$$
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Practice Class 11 Engineering Maths Quiz Questions and Answers
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