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CBSE Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 9 - MCQExams.com
CBSE
Class 11 Engineering Maths
Permutations And Combinations
Quiz 9
The number of such numbers which are even (all digits are different) is
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0%
$$60$$
0%
$$96$$
0%
$$120$$
0%
$$204$$
$$\sum _{ k=1 }^{ 10 }{ k.k! } =$$
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0%
10!
0%
11!
0%
10!+1
0%
11!-1
Explanation
Given,
$$\sum _{k=1}^{10}kk!$$
$$a_k=kk!$$
$$a_1=1\cdot \:1!=1$$
$$a_2=2\cdot \:2!=4$$
Similarly,
$$a_3=18$$
$$a_4=96$$
$$a_5=600$$
$$a_6=4320$$
$$a_7=35280$$
$$a_8=322560$$
$$a_9=3265920$$
$$a_{10}=36288000$$
$$=1+4+18+96+600+4320+35280+322560+3265920+36288000$$
$$=39916799$$
$$=11!-1$$
Solve:
$$\dfrac{^{n}C_{r}}{^{n}C_{r-1}}=$$
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0%
$$\dfrac{n-r}{r}$$
0%
$$\dfrac{n+r-1}{r}$$
0%
$$\dfrac{n-r+1}{r}$$
0%
$$\dfrac{n-r-1}{r}$$
A double Decker is can accommodate 20 passengers 7 in the lower deck 13 in the upper deck. The number of ways the passengers can be accommodate if 5 want to sit only in lower deck and 8 want to sit only in upper deck is
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0%
$$^{7}C_{5}$$
0%
$$^{7}C_{3}$$
0%
$$^{7}C_{1}$$
0%
$$^{7}C_{6}$$
If $$^{n}C_{4},\ ^{n}C_{5}$$ and $$^{n}C_{6}$$ in A.P., then possible value of $$n$$ is
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0%
$$6$$
0%
$$12$$
0%
$$14$$
0%
$$21$$
If $$CARPET$$ is coded as $$TCEAPR$$ then the code for $$NATIONAL$$ would be written as
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0%
$$NLATNOIA$$
0%
$$LANOITAN$$
0%
$$LNAANTOI$$
0%
$$LNOINTAA$$
How many 10 digits number can be written by using digits (9 and 2) ?
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0%
$$^{10}C_1 + ^9C_2$$
0%
$$2^{10}$$
0%
$$^{10} C_2$$
0%
$$10 !$$
Explanation
Total number of numbers =
2
×
2.....
×
10
2×2.....×10
times
⇒
$$2^{10}$$
Hence (B) is the correct answer.
All possible three digits even numbers which can be formed with the condition that if $$5$$ is one of the digit, then $$7$$ is the next digit is:
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0%
$$5$$
0%
$$325$$
0%
$$345$$
0%
$$365$$
Explanation
An even number can be formed by using one of the digits $$0, 2, 4, 6$$ or $$8$$ at the units place.
$$5$$ can not be at ten's place, since in that case $$7$$ has to be placed at the units place. In that case, the number formed is not an even number.
When $$5$$ is at the hundred's place, then $$7$$ will be at the ten's place. The units place can be filled in 5 ways by using one of the digits $$0, 2, 4, 6$$ or $$8.$$
$$\Rightarrow$$
Number of even numbers in this case $$= 1 \times 1 \times 5 = 5$$
When $$5$$ is not at the hundred's place, so the hundred's place can be filled in $$8$$ ways ($$0$$ and $$5$$ cannot be used at hundred's place).
Ten's place can be filled in $$9$$ ways ($$5$$ cannot be used at ten's place).Units place can be filled in $$5$$ ways ($$0, 2, 4, 6,$$ or $$8$$ can occupy the even place).
$$\Rightarrow$$
Number of even number in this case $$= 8 \times 9 \times 5 = 360$$
Hence, by fundamental principle of addition,
$$\Rightarrow$$ Total number of even number $$= 360 + 5 = 365.$$
$$1+1.1!+2.2!+3.3!+...+n.n!$$ is equal to
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0%
$$n!$$
0%
$$(n-1)!$$
0%
$$(n+1)!$$
0%
$$n$$
Explanation
given,
$$1+1.1!+2.2!+3.3!+......+n.n!$$
$$=1+(2-1).1!+(3-1)!.2!+.......+(n+1-1).n!$$
$$=1+2.1!-1.1!+3.2!-1.2!+...+(n+1)n!-1n!$$
$$=1+2!-1!+3!-2!+4!-3!+......(n+1)!-n!$$
$$=1+(n+1)!-1$$
$$=(n+1)!$$
A committee of $$10$$ is to be formed from $$8$$ women and $$6$$ men. In how many of these committees the women are in majority?
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0%
$$515$$
0%
$$545$$
0%
$$575$$
0%
$$595$$
Explanation
for majority, women should be $$ > 5$$ in no.
$$\therefore $$ selecting the committee $$\Rightarrow $$
$$^{8}C_{6}\times ^{6}C_{4}+^{8}C_{7}\times ^{6}C_{3}+^{8}C_{8}\times ^{6}C_{2}$$
$$=\frac{8\times 7\times 6\times 5}{2\times 1\times 2\times 1}+\frac{8}{1}\times \frac{6\times 5\times 4}{3\times 2\times 1}+\frac{8}{8}\times \frac{6\times 5}{2\times 1}$$
$$= 420+160+15$$
$$=595$$
A shelf contains $$15$$ books, of which $$4$$ are single volume and the others are $$8$$ and $$3$$ volumes respectively. In how many ways can these books be arranged on the shelf so that order of the volumes of same work is maintained $$?$$
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0%
$$4!$$
0%
$$8!$$
0%
$$3!$$
0%
$$4!8!3!3!$$
Explanation
No of books $$15$$
No of ways $$4$$ volumes are arranged are $$4!$$ ways
No of ways $$8$$ volumes are arranged are $$8!$$ ways
No of ways $$3$$ volumes are arranged are $$3!$$ ways
They are arranged in $$3!$$ ways one above the other
So no.of ways is $$4!8!3!3!$$
$$\displaystyle \sum^{n-1}_{r=0}\dfrac {^{n}C_{r}}{^{n}C_{r}+^{n}C_{r+1}}=$$
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0%
$$\dfrac {n}{2}$$
0%
$$\dfrac {n+1}{2}$$
0%
$$(n+1)\dfrac {n}{2}$$
0%
$$\dfrac {n\ (n-1)}{2\ (n+1)}$$
Explanation
$$\displaystyle \sum^{n-1}_{r=0}\dfrac {^{n}C_{r}}{^{n}C_{r}+^{n}C_{r+1}}$$
we know $$^{n}C_{r}+^{n}C_{r-1}=^{n+1}C_{r}$$
$$\therefore \Rightarrow \sum^{n-1}_{r=0}\dfrac {^{n}C_{r}}{^{n+1}C_{r+1}}$$
Expanding
$$\Rightarrow \sum^{n-1}_{r=0}\frac{n!}{(n-r)!\cdot r!}\cdot \frac{(n-r)!\cdot (r+1)!}{(n+1)!}$$
$$\Rightarrow \sum^{n-1}_{r=0}\frac{r+1}{n+1}$$
$$\Rightarrow \frac{1+2+3\cdot \cdot \cdot n}{n+1}$$
$$\Rightarrow \frac{n}{2}$$
If $$a=\,^ { m }C _ { 2 } ,$$ then $$^ { a } C _ { 2 }$$ is equal to
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0%
$$^{m + 1}C _ { 4 }$$
0%
$$ ^{m+2} C _ { 4 }$$
0%
$$3.\, ^ { m + 2 } C _ { 4 }$$
0%
$$3. \,^ { m + 1 }C_4$$
Explanation
$${\textbf{Step-1:Define and give examples of nCr in terms of factorials}}$$
$${\text{Given,}}$$
$$\Rightarrow a=^mC_2$$
$$^aC_2=?$$
$$\Rightarrow a=\dfrac{m!}{(m-2)!2!}$$
$$=\dfrac{m(m-1)}{2!}$$
$$=\dfrac{(\dfrac{m(m-1)}{2})(\dfrac{m(m-1)}{2}-1)}{2}$$
$$=\dfrac{(\dfrac{m(m-1)}{2})(\dfrac{m(m-1)-2}{2})}{2}$$
$$=\dfrac{m(m-1)(m-2)(m+1)}{8}$$
$$=\dfrac{(m-1)(m-2)(m)(m+1)}{8}$$
$$=\dfrac{1(m+1)!}{8(m-3)!}$$
$$=\dfrac{3 (m-1)!}{1.2.3.4(m-3)!}$$
$$^aC_2=$$
$$3. \,^ { m + 1 }C_4$$
$${\textbf{Hence option D is correct}}$$
$$^{ n }{ C }_{ 1 }.2+^{ n }{ C }_{ 2 }.\frac { { 2 }^{ 2 } }{ 3 } +^{ n }{ C }_{ 3 }.\frac { { 2 }^{ 3 } }{ { 3 }^{ 2 } } +......^{ n }{ C }_{ n }.\frac { { 2 }^{ n } }{ { 3 }^{ n-1 } } =$$
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0%
$$\frac { { 3 }^{ n }-{ 2 }^{ n } }{ { 3 }^{ n-1 } } $$
0%
$$\frac { { 3 }^{ n }+{ 2 }^{ n } }{ { 3 }^{ n-1 } } $$
0%
$$\frac { { 5 }^{ n }-{ 3 }^{ n } }{ { 3 }^{ n-1 } } $$
0%
$$\frac { { 3 }^{ n }+{ 5 }^{ n } }{ { 3 }^{ n-1 } } $$
The number of seven letter words that can be formed by using the letters of the word $$SUCCESS$$ th
at the two $$C$$ are together but no two $$S$$ are together is
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0%
$$24$$
0%
$$18$$
0%
$$54$$
0%
none of these
Explanation
using the letters of the word $$SUCCESS$$ that the two $$C$$ are together but no two $$S$$ are together
let two C's be 1 unit
$$\therefore $$ no. of ways $$=\frac{6!}{4!}=30$$
We have to put 1 letter between one S
So, No of ways$$=2\times 3!=12$$
No. of ways=30-12=18
Nine boys and 3 girls are to be seated in 2 vans, each having numbered seats, 3 in front and 4 at back. The number of ways of seating arrangements, if the girls should sit together in a back row on adjacent seats, is
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0%
$$12!$$
0%
$$3\times 11!$$
0%
$$4\times 11!$$
0%
$$3\times 9!$$
Six people are going to sit in a row on a bench. $$A$$ and $$B$$ are adjacent, $$C$$ does not want to sit adjacent to $$D.E$$ and $$F$$ can sit anywhere. Number of ways in which these six people can be seated is
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0%
$$200$$
0%
$$144$$
0%
$$120$$
0%
$$56$$
Explanation
A, B, C, D, E, F
Consider AB as group so we have AB, C, D, E, F.
We have totally $$5$$
No. of ways$$(w_1)=5!\times 2$$
$$=240$$
Let CD are adjacent now AB, CD, E, F
No. of ways$$(w_2)=4!2!2!$$
$$=96$$
Total no. of ways
$$W=w_1-w_2$$
$$=240-96$$
$$=144$$.
How many different words can be formed by jumbling the letters in the word $$MISSISSIPPI$$ in which no two $$S$$ are
adjacent ?
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0%
$$8 \times ^ { 6 } C _ { 4 } \times ^ { 7 } C _ { 4 }$$
0%
$$6 \times 7 \times ^ { 8 } C _ { 4 }$$
0%
$$6 \times 8 \times ^ { 7 } C _ { 4 }$$
0%
$$7\times ^{ { 6 } }{ C }_{ { 4 } }\times ^{ { 8 } }{ C }_{ { 4 } }$$
Explanation
We have the word $$MISSISSIPPI$$
In this word there are $$4I,4S,2P\,and\,1M$$
No two $$S$$ should be together,
Hence, we can place $$S$$ at these places
Therefore, the possible number of words is given by
$$\begin{array}{l} \frac { { ^{ 8 }{ C_{ 4 } }.7! } }{ { 4!2! } } \\ =\frac { { ^{ 8 }{ C_{ 4 } }.7.6! } }{ { 4.2! } } \\ ={ 7.^{ 8 } }{ C_{ 4 } }{ .^{ 6 } }{ C_{ 4 } } \\ Hence,\, option\, D\, is\, the\, correct\, answer. \end{array}$$
In the expansion of $$\left(x^3 - \dfrac{1}{x^2}\right)^{15}$$, the constant terms is
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0%
$$^{15}C_6$$
0%
$$-{^{15}C_6}$$
0%
$$^{15}C_4$$
0%
$$-{^{15}C_4}$$
Explanation
We have,
$${ \left( { { x^{ 2 } }-\dfrac { 1 }{ { { x^{ 2 } } } } } \right) ^{ 15 } } \\ { T_{ r+1 } }{ =^{ 15 } }{ C_{ r } }{ \left( { -1 } \right) ^{ r } }{ x^{ 15-r } } \\ { =^{ 15 } }{ C_{ r } }{ \left( { -1 } \right) ^{ r } }{ x^{ 45-5r } } $$
$$\\ For\, constant\ term\, \\ 45-5t=0 \\ r=9 $$
Therefore,
$$=- ^{ 15 }{ C_{ 9 } } = -^{ 15 }{ C_{ 6 } }$$
Hence, this is the answer.
The value of $$^{47}C_{4}+\displaystyle \sum _{ j=1 }^{ 5 }\ ^{ \left( 52-j \right) } { C }_{ 3 }$$ is
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0%
$$^{47}C_{5}$$
0%
$$^{52}C_{5}$$
0%
$$^{52}C_{4}$$
0%
$$^{52}C_{3}$$
Explanation
We have,
$$\begin{array}{l} ^{ 47 }{ C_{ 4 } }{ +^{ 51 } }{ C_{ 3 } }{ +^{ 50 } }{ C_{ 3 } }{ +^{ 49 } }{ C_{ 3 } }{ +^{ 48 } }{ C_{ 3 } }{ +^{ 47 } }{ C_{ 3 } } \\ { =^{ n } }{ C_{ r } }{ +^{ n } }{ C_{ r-1 } }{ =^{ n+1 } }{ C_{ r } } \\ { =^{ 48 } }{ C_{ 4 } }{ +^{ 48 } }{ C_{ 3 } }{ +^{ 49 } }{ C_{ 3 } }{ +^{ 50 } }{ C_{ 3 } }{ +^{ 51 } }{ C_{ 3 } } \\ { =^{ 49 } }{ C_{ 4 } }{ +^{ 49 } }{ C_{ 3 } }{ +^{ 50 } }{ C_{ 3 } }{ +^{ 51 } }{ C_{ 3 } } \end{array}$$
Similarly,
$$=^{52}{C_4}$$.
A committee of $$4$$ persons is to be formed from $$2$$ ladies, $$2$$ old men and $$4$$ young men such that it includes at least $$1$$ lady. at least $$1$$ old man and at most $$2$$ young men. Then the total number of ways in which this committee can be formed is :
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0%
$$40$$
0%
$$41$$
0%
$$16$$
0%
$$32$$
The number of values of 'r' satisfying the equation, $${^{39}C_{3r-1}}-{^{39}C_{r^2}}={^{39}C_{r^2-1}}-{^{39}C_{3r}}$$ is?
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0%
$$1$$
0%
$$2$$
0%
$$3$$
0%
$$4$$
The number lock of a suitcase has four wheels, each labelled with 10-digits i.e., from 0 toThe lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the right sequence to open the suitcase
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0%
$$\dfrac { 1 }{ 5040 } $$
0%
$$\dfrac { 3 }{ 5040 } $$
0%
$$\dfrac { 7 }{ 5040 } $$
0%
None of these
An old man while dialing a $$7$$ digit telephone number remembers that the first four digits consists of one $$1's$$, one $$2's$$ and two $$3's$$. He also remembers that the fifth digits is either a $$4$$ or $$5$$ while has no memorising of the sixth digit, he remembers that the seventh digit is $$9$$ minus the sixth digit. Maximum number of distinct trials he has to try to make sure that he dials the correct telephone number, is
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0%
$$360$$
0%
$$240$$
0%
$$216$$
0%
None of these
Explanation
The telephone number has $$7$$ digits.
The first $$4$$ digits have one $$1$$, one $$2$$ and $$2$$ $$3$$'s.
$$\therefore$$ No. of ways of arranging the first $$4$$ digits$$=\dfrac{4!}{2!}=12$$ ways
The $$5^{th}$$ digit is a $$4$$ or a $$5$$.
$$\therefore$$ There are $$2$$ options for the $$5^{th}$$ digit.
Let the $$6^{th}$$ digit be x. Then, the $$7^{th}$$ digit is $$9-x$$.
$$\therefore$$ Possible combinations for $$6^{th}$$ and $$7^{th}$$ digits are:
($$6^{th}$$ digit, $$7^{th}$$ digit): $$(0, 9), (1, 8), (2, 7), (3, 6), (4, 5), (5, 4), (6, 3), (7, 2), (8, 1), (9, 0)$$.
$$\therefore$$ There are $$10$$ possible combinations for the $$6^{th}$$ and $$7^{th}$$ digits.
$$\therefore$$ No. of possible telephone numbers$$=12\times 2\times 10=240$$, to dial correct number.
$$\therefore$$ The correct answer is option (B).
The number of different seven digit numbers that can be written using only three digits 1, 2 & 3 under the condition that the digit 2 occurs exactly twice in each number is-
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0%
672
0%
640
0%
512
0%
None of these
Explanation
choose any two of the seven digit
This may be done in $$^{7}C_{2}$$ ways
put 2 in these two digits .
The remaining 5 digit may be arranged
using 1 and 3 in $$2^{5}$$ ways.
so, squired numbers of numbers
=$$^{7}C_{2} \times 2^{5}$$
=$$\dfrac{7!}{5!2!}\times 2^{5}$$
=$$\dfrac{7\times 6\times 5!}{5!\times 2}\times 2^{5}$$
=$$21\times 32=672$$
If $$^{8}C_{r}=^{8}C_{3}$$, then $$r$$ is equal to
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0%
$$5$$
0%
$$4$$
0%
$$8$$
0%
$$6$$
Explanation
$$^8C_r=^8C_3$$
$$\frac {8!}{r!(8-r)!}=\frac{8!}{3!\,5!}$$
$$\Rightarrow r!(8-r)!=3 !5!$$
$$\Rightarrow either \, r=3 \,\,\, or r=5 \Rightarrow (A)$$
Find $$x$$, if $$\dfrac {1}{4!}-\dfrac {1}{x}=\dfrac {1}{5!}$$.
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0%
$$5$$
0%
$$4$$
0%
$$30$$
0%
$$None$$
The value of $$\sum _ { r = 1 } ^ { 5 } r \dfrac { ^ { n } C _ { r } } { ^ { n } C _ { r - 1 } } =?$$
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0%
$$5 ( n - 3 )$$
0%
$$5 ( n - 2 )$$
0%
$$5 \mathrm { n }$$
0%
$$5 ( 2 n - 9 )$$
When $$n!+1$$ is divided by any natural number between $$2$$ and $$n$$ then remainder obtained is
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0%
$$1$$
0%
$$2$$
0%
$$3$$
0%
$$4$$
If $$(1 + x)^n = \displaystyle \sum^{n}_{r = 0} {^nC_r} x^r$$ then $$C^2_0 + \dfrac{C^2_1}{2} + \dfrac{C^2_2}{3} + ... + \dfrac{C^2_n}{n + 1} =$$
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0%
$$\dfrac{{2n}!}{n !)^2}$$
0%
$$\dfrac{{2n + 1}!}{{(n + 1)^2}!}$$
0%
$$\dfrac{{2n - 1}!}{{(n + 1)^2}!}$$
0%
$$\dfrac{{n}!}{{(n - 1)^2}!}$$
Set of value of r for which, $$^{18}C_{r-2}+2\cdot {^{18}C_{r-1}}+{^{18}C_{r}} \geq {^{20}C_{13}}$$ contains?
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0%
$$4$$ elements
0%
$$5$$ elements
0%
$$7$$ elements
0%
$$10$$ elements
The no.of triangles formed by selecting the points from Regular pentagon is
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0%
$$10$$
0%
$$12$$
0%
$$16$$
0%
$$none$$
Explanation
The no .of points in pentagon is $$5$$
The points required to form rectangle is $$3$$
The no .of ways of selecting is $$^5C_3=10$$
$$^{n}C_{r}+2^{n}C_{r+1}+^{n}C_{r+2}$$ is equal to
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0%
$$2.^{n}C_{r+2}$$
0%
$$^{n+1}C_{r+1}$$
0%
$$^{n+2}C_{r+2}$$
0%
$$none\ of\ these$$
Value of $$\displaystyle \sum _{ r=0 }^{ n} r.\left(^{n}C_{r}\right)^{2}$$ is equal to
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0%
$$n.{^{2n}C_{r}}$$
0%
$$\dfrac{n.{^{2n}C_{r}}}{2}$$
0%
$$n^{2}.{^{2n}C_{r}}$$
0%
$$\dfrac{n^{2}.{^{2n}C_{r}}}{2}$$
The no .of ways of selecting a prime numbers from First 10 natural numbers is
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0%
$$^{10}C_4$$
0%
$$^4C_{10}$$
0%
$$^{10}P_4$$
0%
$$^{10}C_5$$
Explanation
The prime numbers are $$2,3,5,7$$
No.of numbers $$=4$$
Total no .of numbers $$10$$
No .of ways to select is given as $$^{10}C_4$$
'
A rectangle with sides 2m - 1 and 2n - 1 is divided into square of unit length by drawing parallel lines as shown in diagram, then the number of rectangles possible with odd side length is
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0%
$$(m+n-1)^2$$
0%
$$4^{m+n-1}$$
0%
$$m^2-n^2$$
0%
$$m(m+1)n(n+1)$$
Explanation
REF.Image
A rectangle with sides 2m-1 and 2n-1 is divide into square units length by drawing parallel lines as shown in diagram, then the number of rectangle possible wide odd side length
Sol:
There are 2m vertical (number 1,2........2m) and 2n horizontal lines (numbered 1,2.......2n).
To form the required rectangle we must selected two horizontal lines, one even numbered and one odd numbered and similarly two vertical lines.
The number of rectangle is then
$$(1+3+5+......+(2m-1))(1+3+5+......+(2n-1))-m^{2}.n^{2}$$
$$\therefore $$ so, the answer is $$c.m^{2}-n^{2}$$
How many integers are there such that $$2 \le n \le 100$$ and the highest common factor of $$n$$ and $$36$$ is $$1$$?
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0%
$$166$$
0%
$$332$$
0%
$$331$$
0%
$$416$$
Explanation
$$36.2^{2}.3^{2}$$
Since HCF (36,n) =1
Therefore, n sholud not be a multiple of 2 or 3.
$$ 2 \leq n\leq 100$$
Total numbers = T= 1000.
Number of numbers divisible by 2= $$N_{2}$$
Number of numbers divisible by 3= $$N_{3}$$
Number of numbers divisible by 6= $$N_{6}$$
Therefore, there are $$T-N_{2}-N_{3}+N_{6}$$ total integers
a=2
l=1000
d=2
We know, $$l=a+(N_{2}-1)d.$$
$$N_{2}=\frac{1000-2}{2}+1$$
$$N_{2}=\frac{998}{2}+1$$
$$N_{2}=499+1=500$$
Similarly,
$$N_{3}=\frac{999-3}{3}+1=333$$
$$N_{6}=\frac{999-6}{3}+1=166$$
Answer
= 999-500-333+166
= 332
If $$^mC_3+^mC_4>^{m+1}C_3$$, then least value of $$m$$ is :
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0%
6
0%
7
0%
5
0%
None of these
Explanation
$$^mC_3+^mC_4>^{m+1}C_3$$
Expanding the given
$$\frac{m!}{(m-3)!\cdot 3!}+\frac{m!}{(m-4)!\cdot 4!}>\frac{(m+1)!}{(m-2)!\cdot 3!}$$
Dividing m! and multiplying $$(m-4)!\cdot 3!$$
$$\frac{1}{(m-3)}+\frac{1}{4}>\frac{m+1}{(m-2)(m-3)}$$
Further solving
$$4(m-2)+(m-2)(m-3)>4(m+1)$$
$$m^2-5m-6>0$$
$$(m-6)(m+1)>0$$
$$\therefore m>6,m>-1$$
m cannot be -ve so, m>6
so answer will be 7
Answer
The number of ways in which $$9$$ persons can be divided into three equal groups, is
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0%
$$1680$$
0%
$$840$$
0%
$$560$$
0%
$$280$$
The value of $$\left( \begin{matrix} 30 \\ 0 \end{matrix} \right) \left( \begin{matrix} 30 \\ 10 \end{matrix} \right) -\left( \begin{matrix} 30 \\ 1 \end{matrix} \right) \left( \begin{matrix} 30 \\ 11 \end{matrix} \right) +\left( \begin{matrix} 30 \\ 2 \end{matrix} \right) \left( \begin{matrix} 30 \\ 12 \end{matrix} \right) .....+\left( \begin{matrix} 30 \\ 20 \end{matrix} \right) \left( \begin{matrix} 30 \\ 30 \end{matrix} \right) $$ is, where $$\left( \begin{matrix} n \\ r \end{matrix} \right) =^{ n }{ C }_{ r }.$$
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0%
$$\left( \begin{matrix} 30 \\ 10 \end{matrix} \right) $$
0%
$$\left( \begin{matrix} 30 \\ 15\end{matrix} \right) $$
0%
$$\left( \begin{matrix} 60 \\ 30\end{matrix} \right) $$
0%
$$\left( \begin{matrix} 31\\ 10 \end{matrix} \right) $$
A school committee consists of $$2$$ teachers and $$4$$ students. The number of different committees that can be formed from $$5$$ teachers and $$10$$ students is
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0%
$$200$$
0%
$$2100$$
0%
$$2000$$
0%
$$3200$$
Explanation
Given, a school committee consists of $$2$$ teachers and $$4$$ students.
The number of different committees that can be formed is simply, choosing $$2$$ out of the $$5$$ teachers and $$4$$ students out of the $$10$$ students, we get -
$$\Rightarrow$$ $$n=\,^5C_2\times \,^{10}C_4$$
$$=\dfrac{5\times 4}{2\times 1}\times\dfrac{10\times 9\times 8\times 7}{4\times 3\times 2\times 1}$$
$$=2100$$
Number of cyphers at the end of $$^{2002}$$ C$$_{1001}$$ is
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0%
$$0$$
0%
$$1$$
0%
$$2$$
0%
None of these
If $$\displaystyle \sum _{ r=0 }^{ n }{ \left\{ \dfrac { { n{ C }_{ r-1 } } }{ n{ C }_{ r }+n{ C }_{ r-1 } } \right\} } =2 $$ then n is equal to
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0%
3
0%
4
0%
5
0%
6
Explanation
Given,
$$\displaystyle \sum \limits_{ r=0 }^{ n }{ \left\{ \frac { { ^n{ C }_{ r-1 } } }{ n{ C }_{ r }+n{ C }_{ r-1 } } \right\} } =2 $$
or,
$$\displaystyle \sum\limits _{ r=0 }^{ n }{ \left\{ \frac { { ^n{ C }_{ r-1 } } }{^{( n+1)}{ C }_{ r } } \right\} } =2 $$
or,
$$\sum\limits _{ r=0 }^{ n }\dfrac{\dfrac{n!}{(r-1)!(n-r+1)!}}{\dfrac{(n+1)!}{(n+1-r)!(r)!}} =2$$
or,
$$\sum\limits _{ r=0 }^{ n }{\dfrac{r}{n+1} } =2$$
or, $$\dfrac{1}{n+1}\times \dfrac{n(n+1)}{2}=\dfrac{}{}$$
or, $$\dfrac{n}{2}=2$$
or, $$n=4$$.
The number of all the possible selection which a student can make for answering one or more questions out of eight given question in a paper, which each question has an alternative is
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0%
$$255$$
0%
$$6560$$
0%
$$6561$$
0%
none of these
Explanation
No. of ways$$={^{8}C_1}+{^{8}C_2}+{^{8}C_3}+{^{8}C_4}+{^{8}C_5}+{^{8}C_6}+{^{8}C_7}+{^{8}C_8}$$
$$=({^{8}C_0}+{^{8}C_1}+......{^{8}C_8})-{^{8}C_0}$$
$$=2^8-1=256-1$$
$$=255$$.
If $$\frac{3^{3 n} \cdot 2^{n}}{108}+\frac{3^{3 n}}{729}+\frac{3^{3 n} \cdot 2^{2 n}}{48}+\frac{2^{3 n} \cdot 3^{3 n}}{64}=37^{3} \cdot 3^{6}
$$
, then find the value of n ?
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0%
2
0%
3
0%
4
0%
5
0%
none of these
$$^{n }{ C }_{ r }+^{ n }{ C }_{ r+1 }$$ is equal to______________.
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0%
$$^{ n }{ C }_{ R+1 }$$
0%
$$^{ n }{ C }_{ R+1 }$$
0%
$$^{ n+1 }{ C }_{ R+1 }$$
0%
$$^{ n-1 }{ C }_{ R+1 }$$
Explanation
$$^nC_{r} + ^nC_{r+1}$$
$$= \dfrac{n!}{r!(n-r)!} + \dfrac{n!}{(r+1)!(n-r-1)!}$$
$$= \dfrac{(r+1)\times n!}{(r+1)!(n-r)!} + \dfrac{(n-r)n!}{(r+1)!(n-r)!}$$
Taking LCM we get,
$$= \dfrac{(n+1)n!}{(r+1)!(n-r)!}$$
$$= \dfrac{(n+1)!}{(r+1)!((n+1)-(r+1))!}$$
$$=\, ^{n+1}C_{r+1}$$
If $$^{n}C_{3} + ^{n}C_{4} > ^{n + 1}C_{3}$$, then
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0%
$$n + 1$$
0%
$$\dfrac {n}{2}$$
0%
$$n + 2$$
0%
None of these
$$\displaystyle\sum^{m}_{r=0}{^{n+r}C_n}$$ is equal to?
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0%
$$^{n+m+1}C_{n+1}$$
0%
$$^{n+m+2}C_n$$
0%
$$^{n+m+3}C_{n-1}$$
0%
None of these
Explanation
since $$^{n}\textrm{C}_r=^{n}\textrm{C}_{n-r}$$
$$^{n}\textrm{C}_{n-r}+^{n}\textrm{C}_{n}=^{n+1}\textrm{C}_{r}$$
$$\sum^{m}_{r=0}{^{n+r}C_n}=\sum^{m}_{r=0}{^{n+r}C_r}=^{n}\textrm{C}_{0}+^{n+1}\textrm{C}_{1}+^{n+2}\textrm{C}_{2} \cdot \cdot \cdot\cdot ^{n+m}\textrm{C}_{m} $$
$$=[1 + (n+1)]+^{n+2}\textrm{C}_{2}+^{n+3}\textrm{C}_{3}\cdot \cdot \cdot \cdot ^{n+m}\textrm{C}_{m}$$
$$=[^{n+2}\textrm{C}_{1}+^{n+2}\textrm{C}_{2}]+^{n+3}\textrm{C}_{3}\cdot \cdot \cdot \cdot ^{n+m}\textrm{C}_{m}$$
$$\because [n+2]=^{n+2}\textrm{C}_{1}$$
$$=[^{n+3}\textrm{C}_{2}+^{n+4}\textrm{C}_{3}]+^{n+4}\textrm{C}_{4}\cdot \cdot \cdot \cdot ^{n+m}\textrm{C}_{m}$$and so on....
$$=^{n+m}\textrm{C}_{m-1}+^{n+m}\textrm{C}_{m-1}$$
$$=^{n+m+1}\textrm{C}_{m}=^{n+m+1}\textrm{C}_{n+1}\Rightarrow [^{n}\textrm{C}_{r}=^{n}\textrm{C}_{n-r}]$$
Required answer
The number of permutations which can be formed out of the letters of the word "SERIES" three letters together, is:
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0%
120
0%
60
0%
42
0%
none
The coefficient of $$x^{18}$$ in the expansion of $$(1+x)(1-x)^{10}\{(1+x+x^2)^9\}$$ is?
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0%
$$84$$
0%
$$126$$
0%
$$-42$$
0%
$$42$$
Explanation
Coefficient of $$x^{18}$$ in $$(1+x)(1-x)^{10}(1+x+x^2)^9$$
$$=$$Coefficient of $$x^{18}$$ in $$(1-x)^2\{(1-x)(1+x+x^2)\}^9$$
$$=$$Coefficient of $$x^{18}$$ in $$(1-x^2)(1-x^3)^9$$
$$={^9C_6}=84$$.
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