If $$ ^{n-1}C_r=(k^2-3 ) ^nC_{r+1} $$, then $$ k\in $$

$$ [-\sqrt { 3 } ,\sqrt { 3 } ] $$

MCQ Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 10

$\alpha = ^{m}C_{2}$ , then

$^{\alpha}C_{2}$ is equal to

0%
$^{m + 1}C_{4}$
0%
$^{m - 1}C_{4}$
0%
$3 \ \ ^{m + 2}C_{4}$
0%
$3 \ \ ^{m + 1}C_{4}$

Explanation

$\alpha = ^{m}C_{2}\Rightarrow \alpha = \dfrac {m(m -1)}{2}$

$\therefore ^{a}C_{2} = \dfrac {\alpha (\alpha - 1)}{2} = \dfrac {1}{2} \dfrac {m(m - 1)}{2} \left \{\dfrac {m(m - 1)}{2} - 1\right \}$

$= \dfrac {1}{8} m(m - 1)(m - 2)(m + 1)$

$= \dfrac {1}{8} (m + 1)m(m -1)(m - 2) = 3 \ \ ^{m + 1}C_{4}$ .

All possible number are formed using the digits

$1,1,2,2,2,2,3,4,4$ taken all at a time. The number of such number in which the odd digits occupy even places is:

0%
$175$
0%
$162$
0%
$160$
0%
$180$

Explanation

Number of such number

$=^4C_3\times \dfrac{3!}{2!}\times \dfrac{6!}{2!4!}=180$
1246613_1614136_ans_4d48346ac0bb4370a5ac657d4b094039.png

There are

$31$ objects in a bag in which

$10$ are identical, then the number of ways of choosing

$10$ objects from bag is?

0%
$2^{20}$
0%
$2^{20}-1$
0%
$2^{20}+1$
0%
$2^{21}$

Explanation

$^{21}C_0+{^{21}C_1}+{^{21}C_2}+.....+{^{21}C_{10}}=\dfrac{2^{21}}{2}=2^{20}$ .

A group of students comprises of

$5$ boys and

$n$ girls. If the number of ways, in which a team of

$3$ students can randomly be selected from this group such that there is at least one boy and at least one girl in each team, is

$1750$ , then

$n$ is equal to

0%
$25$
0%
$28$
0%
$27$
0%
$24$

Explanation

$\textbf{Step 1: Find number of ways of selecting 1 boy , 2 girls and 1 girl , 2 boys.}$

$\text{Given, }$

$\text{5 boys and n girls}$

$\text{Number of ways of selecting a team with 1 boy and 2 girls}=^{5}C_{1}. ^{n}C_{2}$

$\text{Number of ways of selecting a team with 2 boy and 1 girls}=^{5}C_{2} . ^{n}C_{1}$

$^{5}C_{1}\cdot\ ^{n}C_{2} + ^{5}C_{2} \cdot\ ^{n}C_{1} = 1750$

$[\textbf{Given}]$

$5\cdot \dfrac {n(n-1)}{2!}+10\cdot n=1750$

$n^{2} + 3n = 700$

$\textbf{Step 2: Solving quadratic equation}$

$n^2+3n-700=0$

$\Rightarrow n^2+28n-25n -700=0$

$\Rightarrow (n-25)(n+28)=0$

$\Rightarrow n=25 \text{ or } n=-28$

$\therefore n = 25$ .

$\textbf{Hence, Option A is correct.}$

A team of three persons with at least one boy and atleast one girl is to be formed from

$5$ boys and

$n$ girls. If the number of sum teams is

$1750$ , then the value of

$n$ is

0%
$24$
0%
$28$
0%
$27$
0%
$25$

Explanation

Given

$5$ boys,

$n$ girls

$(1B,2G)+(2B,1G)$

${ _{ }^{ 5 }{ C } }_{ 1 }.{ _{ }^{ n }{ C } }_{ 2 }+{ _{ }^{ 5 }{ C } }_{ 2 }.{ _{ }^{ n }{ C } }_{ 1 }=1750\Rightarrow 5.\cfrac { n(n-1) }{ 2 } +10.n=1750\Rightarrow \cfrac { n(n-1) }{ 2 } +2n=350\Rightarrow { n }^{ 2 }-n+4n=700$

${ n }^{ 2 }+3n-700=0\Rightarrow (n+28)(n-25)=0\Rightarrow n=25,-28$

$\neq -28$ as number of teams cannot be negative

$a, b$ and

$c$ are the gratest values of

$^{19}C_p, \, ^{20}C_q$ and

$^{21}C_r$ respectively, then :

0%
$\dfrac{a}{10} = \dfrac{b}{11} = \dfrac{c}{21}$
0%
$\dfrac{a}{10} = \dfrac{b}{11} = \dfrac{c}{42}$
0%
$\dfrac{a}{11} = \dfrac{b}{22} = \dfrac{c}{21}$
0%
$\dfrac{a}{11} = \dfrac{b}{22} = \dfrac{c}{42}$

Explanation

We know that

$^nC_r$ is maximum when

$𝑟 =\dfrac n2$ where

$n$ is even

and

$r=\dfrac{n+1}2 \ or \ \dfrac{n-1}2$ when

$n$ is odd

$^nCr$ is maximum at middle term

$a =\, ^{19}C_p = \,^{19}C_{10} =\, ^{19}C_9$

$b = \, ^{20}C_q = \,^{20}C_{10}$

$c = \, ^{21}C_r = \,^{21}C_{10} = \, ^{21}C_{11}$

$\dfrac{a}{^{19}C_9} = \dfrac{b}{\dfrac{20}{10}\times\, ^{19}C_{9}} = \dfrac{c}{\dfrac{21}{11}\times \dfrac{20}{10}\times \,^{19}C_9}$

$\Rightarrow \dfrac{a}{1} = \dfrac{b}{2} = \dfrac{c}{\dfrac{42}{11}}$

$\Rightarrow \dfrac{a}{11} = \dfrac{b}{22} = \dfrac{c}{42}$

$\therefore$

$\boxed{\dfrac{a}{11} = \dfrac{b}{22} = \dfrac{c}{42}}....Answer$

Hence option

$'D'$ is the answer.

$\dfrac{1}{6!}+\dfrac{1}{7!}=\dfrac{x}{8!}$ , then

$x=?$

0%
$32$
0%
$48$
0%
$56$
0%
$64$

Explanation

$\dfrac{1}{6!}+\dfrac{1}{7!}=\dfrac{x}{8!}\Rightarrow \dfrac{8\times 7}{8\times 7\times (6!)}+\dfrac{8}{8\times (7!)}=\dfrac{x}{8!}$

$\Rightarrow \dfrac{56}{8!}+\dfrac{8}{8!}=\dfrac{x}{8!}\Rightarrow x=56+8=64$ .

$\dfrac{^nC_r}{^nC_{r-1}}=?$

0%
$\dfrac{n-r}{r}$
0%
$\dfrac{n-r-1}{r}$
0%
$\dfrac{n-r+1}{r}$
0%
None of these

Explanation

$\dfrac{^nC_r}{^nC_{r-1}}\\=\dfrac{n!}{(r!)\times (n-r)!}\times \dfrac{(r-1)!\times (n-r+1)!}{n!}\\$

$=\dfrac{(r-1)!\times (n-r+1)\times (n-r)!}{r\cdot (r-1)!\times (n-r)!}\\=\dfrac{(n-r+1)}{r}$ .

$^{36}C_{34}=?$

0%
$1224$
0%
$612$
0%
$630$
0%
None of these

Explanation

$^{36}C_{34}={^{36}C_{(36-34)}}={^{36}C_2}=\dfrac{36\times 35}{2}=630$ .

There are

$10$ points in a plane, out of which

$4$ points are collinear. The number of line segments obtained from the pairs of these points is?

0%
$39$
0%
$40$
0%
$41$
0%
$45$

Explanation

Number of line segments formed by joining pairs of points out of

$10$

$={^{10}C_2}=\dfrac{10\times 9}{2}=45$ .

Number of line segments formed by joining pairs of

$4$ points

$={^4C_2}=\dfrac{4\times 3}{2}=6$ .

But, these points being collinear give only one line.

$\therefore$ required number of line segments

$=(45-6+1)=40$ .

$^{n}C_3=220$ , then

$n=?$

0%
$9$
0%
$10$
0%
$11$
0%
$12$

Explanation

$^nC_3=220\Rightarrow \dfrac{n(n-1)(n-2)}{6}=220$

$\Rightarrow n(n-1)(n-2)=1320$

$\Rightarrow n=12$

$[\because 12\times 11\times 10=1320]$ .

$^nC_{10}={^{n}C_{14}}$ , then

$n=?$

0%
$4$
0%
$24$
0%
$14$
0%
$10$

Explanation

$^nC_p={^nC_q}\Rightarrow p+q=n$ .

$\therefore {^nC_{10}}={^{n}C_{14}}\Rightarrow n=(10+14)=24$ .

There are

$10$ points in a plane, out of which

$4$ points are collinear. The number of triangles formed with vertices as these point is?

0%
$20$
0%
$120$
0%
$116$
0%
None of these

Explanation

Number of triangles obtained from

$10$ points

$={^{10}C_3}=\dfrac{10\times 9\times 8}{3\times 2\times 1}=120$ .

Number of triangles obtained from

$4$ points

$={^4C_3}={^4C_1}=4$

But, these

$4$ points being collinear will give no triangle.

$\therefore$ required number of triangles

$=(120-4)=116$ .

$^{n}C_r+{^nC_{r+1}}={^{n+1}C_x}$ , then

$x=?$

0%
$r-1$
0%
$r$
0%
$r+1$
0%
$n$

Explanation

We know that

$^{n}C_r+{^nC_{r+1}}={^{n+1}C_{r+1}}$ .

So,

$x=(r+1)$ .

$^{60}C_{60}=?$

0%
$60!$
0%
$1$
0%
$\dfrac{1}{60}$
0%
None of these

Explanation

$^{60}C_{60}=1$

$[\because {^nC_n}=1]$ .

Out of

$7$ consonants and

$4$ vowels, how many words of

$3$ consonants and

$2$ vowels can be formed?

0%
$330$
0%
$1050$
0%
$6300$
0%
$25200$

Explanation

Number of ways of selecting

$3$ consonants out of

$7$ and

$2$ vowels out of

$4$

$=(^7C_3\times {^4C_2})=\left(\dfrac{7\times 6\times 5}{3\times 2\times 1}\times \dfrac{4\times 3}{2\times 1}\right)=210$ .

Now,

$5$ letters can be arranged among themselves in

$5!$ ways

$=120$ ways.

Required number of words

$=(210\times 120)=25200$ .

$^{n}C_{18}={^nC_{12}}$ , then

$^{32}C_n=?$

0%
$248$
0%
$496$
0%
$992$
0%
None of these

Explanation

$^nC_p={^nC_q}\Rightarrow p+q=n$ .

$^nC_{18}={^{n}C_{12}}\Rightarrow n=(18+12)=30$ .

$\therefore {^{32}C_n}={^{32}C_{30}}={^{32}C_2}=\dfrac{32\times 31}{2}=496$ .

If a denotes the number of permutation of

$x+2$ things taken all at a time,

$b$ the number of permutation of x things taken 11 at a time and c the number of permutation of

$x-11$ things taken all at a time such that

$a=182 bc$ , then the value of

$x$ is

0%
$15$
0%
$12$
0%
$10$
0%
$18$

How many

$3$ -digit numbers are there?

0%
$648$
0%
$729$
0%
$900$
0%
$1000$

Explanation

The hundreds place can be filled by any of the

$9$ non zero digits.

So, there are

$9$ ways of filling this place. The tens place can be filled by any of the

$10$ digits. So, there are

$10$ ways of filling it.

The units place can be filled by any of the

$10$ digits. So, there are

$10$ ways of filling it.

$\therefore$ total number of

$3$ -digit numbers

$=(9\times 10\times 10)=900$ .

When simplified, the expression

$^{ 47 }{ C }_{ 4 }+\sum _{ j=1 }^{ 5 }$

$^{ 52-j }{ C }_{ 3 }$ equals to

0%
$^{ 47 }{ C }_{ 5 }$
0%
$^{ 49 }{ C }_{ 4 }$
0%
$^{ 52 }{ C }_{ 5 }$
0%
$^{ 52 }{ C }_{ 4 }$

The total number of 5- digit telephone numbers that can be composed with distinct digits, is

0%
$^{10}P_{2}$
0%
$^{10}P_{5}$
0%
$^{10}C_{2}$
0%
None of these

$^{ n }{ C }_{ r-1 }=10,$

$^{ n }{ C }_{ r }=45,$ and

$^{ n }{ C }_{ r+1 }=120,$ then r equals

0%
$1$
0%
$2$
0%
$3$
0%
$4$

In an examination, a candidate has to pass in each of the five subjects. In how many ways can he fail?

0%
$5$
0%
$10$
0%
$21$
0%
$31$

Explanation

The candidate can fail by failing in

$1$ or

$2$ or

$3$ or

$4$ or

$5$ subjects out of

$5$ in each case.

$\therefore$ required number of ways

$={^5C_1}+{^5C_2}+{^5C_3}+{^5C_4}+{^5C_5}$

$={^5C_1}+{^5C_2}+{^5C_{(5-3)}}+{^5C_{(5-4)}}+1$

$={^5C_1}+{^5C_2}+{^5C_2}+{^5C_1}+1$

$=2({^5C_1}+{^5C_2})+1=2\left(5+\dfrac{5\times 4}{2\times 1}\right)+1=(30+1)=31$ .

A committee of

$5$ is to be formed out of

$6$ gents and

$4$ ladies. In how many ways can this be done when each committee may have at the most

$2$ ladies?

0%
$120$
0%
$160$
0%
$180$
0%
$186$

Explanation

We may have:

(i) (

$1$ lad out of

$4$ ) and (

$4$ gents out of

$6$ )

or (ii) (

$2$ ladies out of

$4$ ) and (

$3$ gents out of

$6$ ).

$\therefore$ required number of ways

$=(^4C_1\times {^6C_4})+(^4C_2\times {^6C_3})\\=(4\times {^6C_2})+\left(\dfrac{4\times 3}{2\times 1}\times \dfrac{6\times 5\times 4}{3\times 2\times 1}\right)$

$=\left(4\times \dfrac{6\times 5}{2\times 1}\right)+(6\times 20)\\=(60+120)=180$ .

$12$ persons meet in a room and each shakes hands with all the others. How many handshakes are there?

0%
$144$
0%
$132$
0%
$72$
0%
$66$

Explanation

Number of handshakes

$={^{12}C_2}=\dfrac{12\times 11}{2}=66$ .

In how many ways can a committee of

$5$ members be selected from

$6$ men and

$5$ ladies, consisting of

$3$ men and

$2$ ladies?

0%
$25$
0%
$50$
0%
$100$
0%
$200$

Explanation

Number of ways of selecting

$3$ men out of

$6$ and

$2$ ladies out of

$5$

$=(^6C_3\times {^5C_2})=\left(\dfrac{6\times 5\times 4}{3\times 2\times 1}\times \dfrac{5\times 4}{2\times 1}\right)=200$ .

Out of

$5$ men and

$2$ women, a committee of

$3$ is to be formed. In how many ways can it be formed if at least one woman is included in each committee?

0%
$21$
0%
$25$
0%
$32$
0%
$50$

Explanation

We may have:

(i)

$1$ woman and

$2$ men or (ii)

$2$ women and

$1$ man.

$\therefore$ required number of ways

$=(^2C_1\times {^5C_2})+(^2C_2\times {^5C_1})=\left(2\times \dfrac{5\times 4}{2\times 1}\right)+(1\times 5)$

$=(20+5)=25$ .

The exponent of 3 in

$100!$ is

0%
$12$
0%
$24$
0%
$48$
0%
$96$

The letters of word 'ZENITH' are written in all positive ways. If all these words are written in the order of a dictionary, then the rank of the word 'ZENITH' is

0%
$716$
0%
$692$
0%
$698$
0%
$616$

Explanation

The total number of words is

$6! = 720$ . Let us write the letters of word ZENITH alphabetically, i.e, EHINTZ.

For ZENITH word starts with	Word starting with	Number of words
$Z$	$E$	$5!$
	$H$	$5!$
	$I$	$5!$
	$N$	$5!$
	$T$	$5!$
ZEN	ZEH	$3!$
	ZEI	$3!$
ZENI	ZENH	$2$
ZENIT	ZENIH	$1$
	Total number of words before ZERNITH	$615$

Hence, there are

$615$ words before ZENITH, so the rank of ZENITH is

$616$ .

Find the values of

$^{61}C_{57} -^{60}C_{56}$

0%
$^{61}C_{58}$
0%
$^{60}C_{57}$
0%
$^{60}C^{58}$
0%
$^{60}C_{56}$

Explanation

$^{61}C_{57} -^{60}C_{56} = ^{60+1}C_{57} -^{60}C_{56}$

$=(^{60}C_{57} + ^{60}C_{56}) - ^{60}C_{56}$

$( \because ^{n+1}C_r = ^nC_r + ^nC_{r-1})$

$=^{60}C_{57} + ^{60}C_{56} -^{60}C_{56}$

$=^{60}C_{57}$
hence option (B) correct.

$^{15}C_{3r} = ^{15}C_{r+3}$ then r equal to :

0%
$5$
0%
$4$
0%
$3$
0%
$2$

Explanation

$^{15}C_{3r} =^{15}C_{r+3}$

$\Rightarrow ^{15}C_{3r} =^{15}C_{15-(r+3) } ( \because ^nC_r = ^nC_{n-r})$

$\Rightarrow ^{15}C_{3r} =^{15}C_{12-r}$
On comparing

$3r = 12 -r$

$\Rightarrow 3r +r = 12$

$\Rightarrow 4r = 12$

$\Rightarrow r = 3$
hence option (C) is correct.

$^{ 47 }C_{ 4 }$ +

$\sum _{ r=1 }^{ 5 }$ .

$^{ 52-r }C_{ 3 }$ is equal to :

0%
$^{51}C_4$
0%
$^{52}C_4$
0%
$^{53}C_4$
0%
None of these

Explanation

$= ^{47}C_4+ ^{ 52-1}C_3 + ^{52-2}C_3 + ^{52-3}C_3 + ^{52-4}C_3 +^{52-5}C_3$

$=^{47}C_4 + ^{51}C_3 + ^{50}C_3 + ^{49}C_3 +^{4}C_3 + ^{47}C_3$

$=(^{47}C_4 + ^{47}C_3) + ^{4}C_3 +^{49}C_3 +^{50}C_3 +^{51}C_3$

$= (^{48}C_4 +^{48}C_3 + ^{49}C_3 +^{50}C_3 +^{51}C_3$

$( \because ^nC_r + ^nC_{r-1} = ^{n+1}C_r)$

$= (^{49}C_4 +^{49}C_3) + ^{50}C_3 + ^{51}C_3)( \because ^nC_r + ^nC_{r-1} = ^{n+1}C_r)$

$=(^{50}C_4 + ^{50}C_3) + ^{51}C_3 )(\because ^nC_r +^nC_{r-1} = ^{n+1}C_r)$

$=(^{51}C_4 + ^{51}C_3 \quad ( \because ^nC_r +^nC_{r-1} = ^{n+1}C_r )$

$=^{52}C_4 \quad ( \because ^nC_r + ^nC_{r-1} = ^{n+1}C_r )$

$_{ }^{ n+1 }{ { C }_{ r+1 } }$ :

$_{ }^{ n }{ { C }_{ r } }$ :

$_{ }^{ n-1 }{ { C }_{ r-1 } }=$ 11: 6: 3 , then

$r=$

0%
20
0%
30
0%
40
0%
5

Explanation

$\dfrac{\:^{n+1}C_{r+1}}{\:^{n}C_{r}}=\dfrac{11}{6}$

$\dfrac{(n+1)!.(n-r)!.r!}{(n-r)!(r+1)!n!}=\dfrac{11}{6}$

$\dfrac{n+1}{r+1}=\dfrac{11}{6}$

$6n+6=11r+11$

$6n-11r=5$ ...(i)
and

$\dfrac{\:^{n}C_{r}}{\:^{n-1}C_{r-1}}=2$

$\dfrac{n!(n-r)!(r-1)!}{(n-1)!(n-r)!r!}=2$

$\dfrac{n}{r}=2$

$n=2r$ ...(ii)
Substituting in (i), we get

$6(2r)-11r=5$

$r=5$ and

$n=10$

$\displaystyle x,y\in(0,30)$ such that

$[ \dfrac{x}{3}]+[\dfrac{3x}{2}]+[\dfrac{y}{2}]+[\dfrac{3y}{4}]=\dfrac{11x}{6}+\dfrac{5y}{4}$ (where [x] denote greatest integer

$\le x$ ) then the number of ordered pairs

$(x, y)$ is

0%
10
0%
20
0%
24
0%
28

Explanation

Let

$\{\mathrm{x}\}=\mathrm{x}-[\mathrm{x}]$ denote fractional part of

$\mathrm{x} i.e 0\leq$

$\{\mathrm{x}\}<1$
Now given expression is

$\displaystyle \frac{x}{3}-\{\frac{x}{3}\}+\frac{3x}{2}-\{\frac{3x}{2}\}+\frac{y}{2}-\{\frac{y}{2}\}+\frac{3y}{4}-\{\frac{3y}{4}\}=\frac{11x}{6}+\frac{5y}{4}$

$\Rightarrow \displaystyle \{\frac{x}{3}\}+\{\frac{3x}{2}\}+\{\frac{y}{2}\}+\{\frac{3y}{4}\}=0$

$\therefore \displaystyle \{\frac{x}{3}\}=\{\frac{3x}{2}\}=\{\frac{y}{2}\}=\{\frac{3y}{4}\}=0$

$\Rightarrow x =6,12,18,24$ ;
and

$y =4,8,12,16,20,24,28$
Hence no. of ordered pairs

$=4\times7=28$

$n = ^mC_2$ , then the value of

$^nC_2$ is given by

0%
$^{m+1}C_4$
0%
$^{m-1}C_4$
0%
$^{m+2}C_4$
0%
$3.^{m+1}C_4$

Explanation

Since,

$\displaystyle n = {}^m{C_2} = \frac{{m!}}{{2!\left( {m - 2} \right)!}} = \frac{{m\left( {m - 1} \right)}}{2}$ .
and

$\displaystyle {}^n{C_2} = \frac{{n!}}{{2!\left( {n - 2} \right)!}} = \frac{{n\left( {n - 1} \right)}}{2}$

$\displaystyle \therefore {}^n{C_2} = \frac{{\left( {\frac{{m\left( {m - 1} \right)}}{2}} \right)\left( {\frac{{m\left( {m - 1} \right)}}{2} - 1} \right)}}{2}$

$\displaystyle = \left( {\frac{{m\left( {m - 1} \right)}}{4}} \right)\left( {\frac{{m\left( {m - 1} \right)}}{2} - 1} \right)$

$\displaystyle =\left( {\frac{{m\left( {m - 1} \right)}}{4}} \right)\left( {\frac{{m\left( {m - 1} \right) - 2}}{2}} \right)$

$\displaystyle =\left( {\frac{{m\left( {m - 1} \right)}}{4}} \right)\left( {\frac{{{m^2} - m - 2}}{2}} \right)$

$\displaystyle =\left( {\frac{{m\left( {m - 1} \right)}}{4}} \right)\left( {\frac{{\left( {m - 2} \right)\left( {m + 1} \right)}}{2}} \right)$

$\displaystyle ={\frac{{\left( {m + 1} \right)m\left( {m - 1} \right)\left( {m - 2} \right)}}{{2 \cdot 4}}}$

$\displaystyle =3\frac{{\left( {m + 1} \right)m\left( {m - 1} \right)\left( {m - 2} \right)\left( {\left( {m - 3} \right)!} \right)}}{{\left( {1 \cdot 2 \cdot 3 \cdot 4} \right)\left( {\left( {m - 3} \right)!} \right)}}$

$\displaystyle =3\left( {{}^{m + 1}{C_4}} \right)$

Let

$\displaystyle \mathrm{S}_{1}=\sum_{\mathrm{j}=1}^{10}j(j-1)^{10}\mathrm{C}_{\mathrm{j}}$ ,

$\displaystyle \mathrm{S}_{2}=\sum_{\mathrm{j}=1}^{10}j^{10}\mathrm{C}_{\mathrm{j}}$ and

$\displaystyle \mathrm{S}_{3}=\sum_{\mathrm{j}=1}^{10}j^{2}10_{\mathrm{C}_{\mathrm{j}}}$ .
Statement-1:

$\mathrm{S}_{3}=55\times 2^{9}$
Statement-2:

$\mathrm{S}_{1}=90\times 2^{8}$ and

$\mathrm{S}_{2}=10\times 2^{8}$ .

0%
Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1
0%
Statement-1 is true, Statement-2 is false
0%
Statement-1 is false, Statement-2 is true
0%
Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1

Explanation

$\mathrm{S}_{1}=\sum_{\mathrm{j}=1}^{10}|(j-1)\dfrac{10!}{(j-1)(j-2)!(10-j)!}=90\sum_{\mathrm{j}=2}^{10}\dfrac{8!}{(j-2)!(8-(j-2))!}=90\cdot 2^{8}$

$\mathrm{S}_{2}=\sum_{\mathrm{j}=1}^{10}|\dfrac{10!}{j(j-1)!(9-(j-1))!}=10\sum_{\mathrm{j}=1}^{10}\dfrac{9!}{(j-1)!(9-(j-1))!}=10\cdot 2^{9}$

$\displaystyle \mathrm{S}_{3}=\sum_{\mathrm{j}=1}^{10}[j(j-1) +j]\displaystyle \dfrac{10!}{|!(10-j)!}=\sum_{\mathrm{j}=1}^{10}|(j-1)^{10}\mathrm{C}_{\mathrm{j}}=\sum_{\mathrm{j}=1}^{10}j^{10}\mathrm{C}_{\mathrm{j}}=90.2^{8}+10.2^{9}$

$=90.2^{8}+20.2^{8}=110.2^{8}=55.2^{9}$ .

$^{n-1}C_r=(k^2-3 ) ^nC_{r+1}$ , then

$k\in$

0%
$(\infty ,-2)$
0%
$(2,\infty )$
0%
$[-\sqrt { 3 } ,\sqrt { 3 } ]$
0%
$(\sqrt { 3 } ,2]$

Explanation

$k^{2}-3=\displaystyle \frac{{}^{n-1}C_{r}}{{}^n{C_{r+1}}}$

$=\displaystyle \frac{(n-1)!}{r!.(n-r-1)!}\times\frac{(r+1)!(n-r-1)!}{n!}$

$=\displaystyle \frac{r+1}{n}$

$\therefore k^{2}=\displaystyle \frac{r+1}{n}+3$
Also

$n-1\geq r\Rightarrow n\geq r +1(r \geq 0)$

$k^{2}\in(3,4]\Rightarrow k\in(\sqrt{3},2]$

In a test there were

$n$ questions. In the test

$2^{n - i}$ students gave wrong answers to at least

$i$ questions

$i = 1, 2, 3 .... n$ . If the total number of wrong answers given is

$2047$ , then

$n$ is

0%
$12$
0%
$11$
0%
$10$
0%
$13$

Explanation

Let the number of students who gave wrong answer to exactly

$i$ questions be

$E_i$

Number students who gave wrong answer to at least

$1$ question

$=2^{n-1} = E_1 + E_2 + E_3 + .... E_n$
Number students who gave wrong answer to at least

$2$ questions

$=2^{n-2} = E_2 + E_3 + E_4 + ... + E_n$
Number students who gave wrong answer to at least

$3$ questions

$=2^{n-3} = E_3 + E_4 + E_5 + ... E_n$
:
:
Number students who gave wrong answer to at least

$n$ questions

$=1 = E_n$
Number of wrong answers given

$=E_1 + 2E_2 + 3E_3 + ...+nE_n$
Adding

$n$ equations listed above we have,

$E_1 + 2E_2 + 3E_3 + ...+nE_n =1+2+2^{2}+2^{3}+...2^{n-1}$
Hence,
Sum

$=1+2+2^{2}+2^{3}+...2^{n-1}$

$2047=\dfrac{2^{n}-1}{2-1}$

$2047=2^{n}-1$

$\therefore 2^{n}=2048$

$\therefore n=11$

Which of the following is equal to

$\dfrac{1.3.5....(2n-1)}{2.4.6....(2n)}$ ?

0%
$(2n!) \div (2^n(n!))^2$
0%
$(2n!) \div n!$
0%
$(2n - 1) \div (n - 1)!$
0%
$2^n$

Explanation

$\dfrac { 1.3.5...(2n-1) }{ 2.4.6...(2n) } \\ \Longrightarrow \dfrac { 1.2.3.4...(2n-1).(2n) }{ { (2.4.6...(2n)) }^{ 2 } } \\ \Longrightarrow \dfrac { 2n! }{ { ((2.1)(2.2).(2.3)...(2.n)) }^{ 2 } } \\ \Longrightarrow \dfrac { 2n! }{ { \left( { 2 }^{ n }n! \right) }^{ 2 } } \\ \Longrightarrow \dfrac { 2n! }{ { \left( { 2 }^{ 2n })(n! \right) }^{ 2 } }$

The value of

$x$ in the equation

$3 \times ^{x+1}C_2 = 2\times ^{x+2}C_2, \space x\space \in N$ is

0%
$x = 4$
0%
$x = 5$
0%
$x= 6$
0%
$x= 7$

Explanation

Given:

$3\times \left( \begin{matrix} x+1 \\ 2 \end{matrix} \right) =2\times \left( \begin{matrix} x+2 \\ 2 \end{matrix} \right) \\ \Rightarrow \dfrac { \left( \begin{matrix} x+2 \\ 2 \end{matrix} \right) }{ \left( \begin{matrix} x+1 \\ 2 \end{matrix} \right) } =\dfrac { 3 }{ 2 } \quad \Rightarrow \dfrac { \left( x+2 \right) \left( x+1 \right) }{ \left( x+1 \right) x } =\dfrac { 3 }{ 2 } \\ \quad \Rightarrow 3x=2x+4\quad \quad \Rightarrow x=4$

The value of

$\displaystyle \sum_{r=0}^{n-1} {\;}^nC_r / (^nC_r+^nC_{r+1})$ equals

0%
$n+1$
0%
$n/2$
0%
$n+2$
0%
none of these

Explanation

$\displaystyle \sum_{r=0}^{n-1}\dfrac {^nC_r}{^nC_r+^nC_{r+1}}$

$=\displaystyle \sum_{r=0}^{n-1}\dfrac {1}{1+\dfrac {^nC_{r+1}}{^nC_r}}$

$=\displaystyle \sum_{r=0}^{n-1}\dfrac {1}{1+\dfrac {n-r}{r+1}}$

$=\displaystyle \sum_{r=0}^{n-1}\dfrac {r+1}{n+1}=\dfrac {1}{n+1}\displaystyle \sum_{r=0}^{n-1}(r+1)$

$=\displaystyle \dfrac {1}{(n+1)}[1+2+....+n]=\dfrac {n}{2}$

How many different signals can be made by hoisting

$6$ differently coloured flags one above the other, when any number of them may be hoisted at once?

0%
1956
0%
1955
0%
1900
0%
1901

Explanation

Here the flags are placed one above the another so, the order is important => we will be using permutations
We can hoist flags

$1$ to

$6$ ,

$=_{1}^{6}\textrm{P}+_{2}^{6}\textrm{P}+_{3}^{6}\textrm{P}+_{4}^{6}\textrm{P}+_{5}^{6}\textrm{P}+_{6}^{6}\textrm{P}=1956$ ways
Hence, option 'A' is correct.

$14X's$ have to be placed in the squares of the above figure such that each row contains at least one

$X$ . In how many ways can this be done?

0%
96
0%
104
0%
112
0%
118

Explanation

Total X's =14
Total Blocks = 16
2 extra blocks,
only possible way is if no X is put either in the first row or the last row
No. of ways = Total ways of Putting X's - 2
=

$^{16}C_{14} -2$
=

$118$

The letters of word

$OUGHT$ are written in all possible orders and these words are written out as in a dictionary. Find the rank of the word

$TOUGH$ in this dictionary.

0%
89
0%
90
0%
91
0%
92

Explanation

Starting with G there are

$4!$
Starting with H there are

$4!$
Starting with O there are

$4!$
Starting with TG there are

$3!$
Starting with TH there are

$3!$
Starting with TOG there are

$2!$
Starting with TOH there are

$2!$
Starting with TOUGH there are

$1!$

The rank of the word TOUGH in this dictionary

$=4!+4!+4!+3!+3!+2!+2!+1!={89}^{th} word$

Hence, option 'A' is correct.

If all the permutations of the letters in the word 'OBJECT' are arranged (and numbered serially) in alphabetical order as in a dictionary, then the

$717^{th}$ word is

0%
TOJECB
0%
TOEJBC
0%
TOCJEB
0%
TOJCBE

Explanation

The order of letters of the word "OBJECT' is B C E J O T
Words starting with:
B can be formed in

$5!=120$
C can be formed in

$5!=120$ .Total = 240
E can be formed in

$5!=120$ . Total = 360
J can be formed in

$5!=120$ . Total = 480
O can be formed in

$5!=120$ . Total = 600
TB can be formed in

$4!=24$ . Total = 624
TC can be formed in

$4!=24$ . Total = 648
TE can be formed in

$4!=24$ . Total = 672
TJ can be formed in

$4!=24$ . Total = 696
TO can be formed in

$4!=24$ . Total = 720
So, we have exceeded 717. We need to go back by 3.
Last word with TO is TOJECB ---- number 720
Before that comes TOJEBC ----- number 719
Before that comes TOJCEB ----- number 718
Before that comes TOJCBE ----- number 717
Hence, (D) is correct.

The value of

$^{47}C_4 + \displaystyle\sum_{r=1}^{5}{^{52-r}C_3}$ =

0%
$^{52}C_2$
0%
$^{52}C_3$
0%
$^{52}C_4$
0%
$^{52}C_5$

Explanation

We know that,

$\displaystyle ^{n}{C}_{r}+{}^{n}{C}_{r-1}$

$\displaystyle =\frac{n!}{r!(n-r)!}+\frac{n!}{(r-1)!(n-r+1)!}$

$\displaystyle =\frac{n!(n-r+1)}{r!(n-r)!(n-r+1)}+\frac{n!r}{(r-1)!(n-r+1)!r}$

$\displaystyle =\frac{n!(n-r+1)}{r!(n-r+1)!}+\frac{n!r}{(n-r+1)!r!}$

$\displaystyle =\frac{n!}{r!(n-r+1)!}(n-r+1+r)$

$\displaystyle =\frac{n!}{r!(n-r+1)!}(n+1)$

$\displaystyle =\frac{(n+1)!}{r!(n-r+1)!}$

$=\displaystyle {}^{n+1}{C}_{r}$

$\Rightarrow \displaystyle {}^{n}{C}_{r}+{}^{n}{C}_{r-1}={}^{n+1}{C}_{r}$
The equation is:

$({}^{47}{C}_{4}+{}^{47}{C}_{3})+{}^{48}{C}_{3}+{}^{49}{C}_{3}+{}^{50}{C}_{3}+{}^{51}{C}_{3}$

$=({}^{48}{C}_{4}+{}^{48}{C}_{3})+{}^{49}{C}_{3}+{}^{50}{C}_{3}+{}^{51}{C}_{3}$

$=({}^{49}{C}_{4}+{}^{49}{C}_{3})+{}^{50}{C}_{3}+{}^{51}{C}_{3}$

$=({}^{50}{C}_{4}+{}^{50}{C}_{3})+{}^{51}{C}_{3}$

$=({}^{51}{C}_{4}+{}^{51}{C}_{3})$

$={}^{52}{C}_{4}$

Hence, option 'C' is correct.

$p$ is a prime number and

$n < p < 2n$ . If

$N=^{2n}C_n$ , then

0%
$p$ divides $N$ completely
0%
$p^2$ divides $N$ completely
0%
$p$ cannot divide $N$
0%
none of these

Explanation

Solution:

$N={}^{2n}C_n=\cfrac{2n!}{n!n!}=\cfrac{(n+1)(n+2)(n+3)........(2n)}{(1).(2).......(n)}$

$p>n$ and

$p<2n, p$ occurs exactly once in the numerator but doesn't occur in the denominator.

It means

$p$ divides

$N$ completely.

Hence, A is the correct option.

The rank of the word

$NUMBER$ obtained, if the letters of the word

$NUMBER$ are written in all possible orders and these words are written out as in a dictionary is

0%
$468$
0%
$469$
0%
$470$
0%
$471$

Explanation

Alphabetically,
Starting with B

$5!$ ways
Starting with E

$5!$ ways
Starting with M

$5!$ ways
Starting with NB

$4!$ ways
Starting with NE

$4!$ ways
Starting with NM

$4!$ ways
Starting with NU

$4!$ ways
Starting with NUB

$3!$ ways
Starting with NUE

$3!$ ways
Starting with NUMBER

$1!$ ways
Therefore, the total will be

$3.5!+4.4!+2.3!+1=469$ ways
Hence, option 'B' is correct.

If the difference of the number of arrangements of three things from a certain number of dissimilar things and the number of selections of the same number of things from them exceeds

$100$ , then the least number of dissimilar things is

0%
$8$
0%
$6$
0%
$5$
0%
$7$

Explanation

Here,

$^nP_3-^nC_3 > 100$

or

$\dfrac {n!}{(n-3)!}-\dfrac {n!}{3!(n-3)!} > 100$

or

$\dfrac {5}{6}n(n-1)(n-2) > 100$

or

$n(n-1)(n-2) > 100$

or

$n(n-1)(n-2) > 6\times 5\times 4$

or

$n=7, 8, .....$

$^nC_r + \displaystyle\sum_{j=0}^{3}{^{n+j}C_{r+1+j}} =$ _________________

0%
$^{n+3}C_{r+3}$
0%
$^{n}C_{r+4}$
0%
$^{n+1}C_{r+4}$
0%
$^{n+4}C_{r+4}$

Explanation

Expanding
We get

$\:^{n}C_{r}+\:^{n}C_{r+1}+\:^{n+1}C_{r+2}+\:^{n+2}C_{r+3}+\:^{n+3}C_{r+4}$

$=\:^{n+1}C_{r+1}+\:^{n+1}C_{r+2}+\:^{n+2}C_{r+3}+\:^{n+3}C_{r+4}$

$=\:^{n+2}C_{r+2}+\:^{n+2}C_{r+3}+\:^{n+3}C_{r+4}$

$=\:^{n+3}C_{r+3}+\:^{n+3}C_{r+4}$

$=\:^{n+4}C_{r+4}$

CBSE Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 10 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 10 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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