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CBSE Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 10 - MCQExams.com
CBSE
Class 11 Engineering Maths
Permutations And Combinations
Quiz 10
If $$\alpha = ^{m}C_{2}$$, then $$^{\alpha}C_{2}$$ is equal to
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$$^{m + 1}C_{4}$$
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$$^{m - 1}C_{4}$$
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$$3 \ \ ^{m + 2}C_{4}$$
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$$3 \ \ ^{m + 1}C_{4}$$
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:
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$$175$$
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$$162$$
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$$160$$
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$$180$$
Explanation
Number of such number $$=^4C_3\times \dfrac{3!}{2!}\times \dfrac{6!}{2!4!}=180$$
There are $$31$$ objects in a bag in which $$10$$ are identical, then the number of ways of choosing $$10$$ objects from bag is?
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$$2^{20}$$
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$$2^{20}-1$$
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$$2^{20}+1$$
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$$2^{21}$$
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
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$$25$$
0%
$$28$$
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$$27$$
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$$24$$
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
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$$24$$
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$$28$$
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$$27$$
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$$25$$
Explanation
Given $$5$$ boys, $$n$$ girls
$$(1B,2G)+(2B,1G)$$
$${ _{ }^{ 5 }{ C } }_{ 1 }.{ _{ }^{ n }{ C } }_{ 2 }+{ _{ }^{ 5 }{ C } }_{ 2 }.{ _{ }^{ n }{ C } }_{ 1 }=1750\Rightarrow 5.\cfrac { n(n-1) }{ 2 } +10.n=1750\Rightarrow \cfrac { n(n-1) }{ 2 } +2n=350\Rightarrow { n }^{ 2 }-n+4n=700$$
$${ n }^{ 2 }+3n-700=0\Rightarrow (n+28)(n-25)=0\Rightarrow n=25,-28$$
n $$\neq -28$$ as number of teams cannot be negative
If $$a, b$$ and $$c$$ are the gratest values of $$^{19}C_p, \, ^{20}C_q$$ and $$^{21}C_r$$ respectively, then :
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$$\dfrac{a}{10} = \dfrac{b}{11} = \dfrac{c}{21}$$
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$$\dfrac{a}{10} = \dfrac{b}{11} = \dfrac{c}{42}$$
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$$\dfrac{a}{11} = \dfrac{b}{22} = \dfrac{c}{21}$$
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$$\dfrac{a}{11} = \dfrac{b}{22} = \dfrac{c}{42}$$
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.
If $$\dfrac{1}{6!}+\dfrac{1}{7!}=\dfrac{x}{8!}$$, then $$x=?$$
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$$32$$
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$$48$$
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$$56$$
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$$64$$
Explanation
$$\dfrac{1}{6!}+\dfrac{1}{7!}=\dfrac{x}{8!}\Rightarrow \dfrac{8\times 7}{8\times 7\times (6!)}+\dfrac{8}{8\times (7!)}=\dfrac{x}{8!}$$
$$\Rightarrow \dfrac{56}{8!}+\dfrac{8}{8!}=\dfrac{x}{8!}\Rightarrow x=56+8=64$$.
$$\dfrac{^nC_r}{^nC_{r-1}}=?$$
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$$\dfrac{n-r}{r}$$
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$$\dfrac{n-r-1}{r}$$
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$$\dfrac{n-r+1}{r}$$
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None of these
Explanation
$$\dfrac{^nC_r}{^nC_{r-1}}\\=\dfrac{n!}{(r!)\times (n-r)!}\times \dfrac{(r-1)!\times (n-r+1)!}{n!}\\$$
$$=\dfrac{(r-1)!\times (n-r+1)\times (n-r)!}{r\cdot (r-1)!\times (n-r)!}\\=\dfrac{(n-r+1)}{r}$$.
$$^{36}C_{34}=?$$
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$$1224$$
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$$612$$
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$$630$$
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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?
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$$39$$
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$$40$$
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$$41$$
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$$45$$
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$$=(45-6+1)=40$$.
If $$^{n}C_3=220$$, then $$n=?$$
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$$9$$
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$$10$$
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$$11$$
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$$12$$
Explanation
$$^nC_3=220\Rightarrow \dfrac{n(n-1)(n-2)}{6}=220$$
$$\Rightarrow n(n-1)(n-2)=1320$$
$$\Rightarrow n=12$$ $$[\because 12\times 11\times 10=1320]$$.
If $$^nC_{10}={^{n}C_{14}}$$, then $$n=?$$
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$$4$$
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$$24$$
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$$14$$
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$$10$$
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?
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$$20$$
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$$120$$
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$$116$$
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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 $$=(120-4)=116$$.
If $$^{n}C_r+{^nC_{r+1}}={^{n+1}C_x}$$, then $$x=?$$
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$$r-1$$
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$$r$$
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$$r+1$$
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$$n$$
Explanation
We know that $$^{n}C_r+{^nC_{r+1}}={^{n+1}C_{r+1}}$$.
So, $$x=(r+1)$$.
$$^{60}C_{60}=?$$
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$$60!$$
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$$1$$
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$$\dfrac{1}{60}$$
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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?
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$$330$$
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$$1050$$
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$$6300$$
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$$25200$$
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$$=(210\times 120)=25200$$.
If $$^{n}C_{18}={^nC_{12}}$$, then $$^{32}C_n=?$$
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$$248$$
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$$496$$
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$$992$$
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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
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$$15$$
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$$12$$
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$$10$$
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$$18$$
How many $$3$$-digit numbers are there?
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$$648$$
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$$729$$
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$$900$$
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$$1000$$
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
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$$^{ 47 }{ C }_{ 5 }$$
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$$^{ 49 }{ C }_{ 4 }$$
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$$^{ 52 }{ C }_{ 5 }$$
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$$^{ 52 }{ C }_{ 4 }$$
The total number of 5- digit telephone numbers that can be composed with distinct digits, is
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$$ ^{10}P_{2}$$
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$$ ^{10}P_{5}$$
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$$ ^{10}C_{2}$$
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None of these
If $$^{ n }{ C }_{ r-1 }=10,$$ $$^{ n }{ C }_{ r }=45,$$ and $$^{ n }{ C }_{ r+1 }=120,$$ then r equals
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$$1$$
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$$2$$
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$$3$$
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$$4$$
In an examination, a candidate has to pass in each of the five subjects. In how many ways can he fail?
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$$5$$
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$$10$$
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$$21$$
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$$31$$
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?
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$$120$$
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$$160$$
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$$180$$
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$$186$$
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?
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$$144$$
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$$132$$
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$$72$$
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$$66$$
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?
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$$25$$
0%
$$50$$
0%
$$100$$
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$$200$$
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?
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$$21$$
0%
$$25$$
0%
$$32$$
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$$50$$
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
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0%
$$12$$
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$$24$$
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$$48$$
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$$96$$
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
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$$716$$
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$$692$$
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$$698$$
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$$616$$
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 $$ ^{61}C_{57} -^{60}C_{56} $$
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$$ ^{61}C_{58} $$
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$$ ^{60}C_{57} $$
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$$ ^{60}C^{58} $$
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$$ ^{60}C_{56} $$
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.
If $$ ^{15}C_{3r} = ^{15}C_{r+3} $$ then r equal to :
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$$ 5 $$
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$$ 4 $$
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$$ 3 $$
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$$ 2 $$
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 :
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$$ ^{51}C_4 $$
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$$ ^{52}C_4 $$
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$$ ^{53}C_4 $$
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None of these
Explanation
$$ = ^{47}C_4+ ^{ 52-1}C_3 + ^{52-2}C_3 + ^{52-3}C_3 + ^{52-4}C_3 +^{52-5}C_3 $$
$$ =^{47}C_4 + ^{51}C_3 + ^{50}C_3 + ^{49}C_3 +^{4}C_3 + ^{47}C_3 $$
$$ =(^{47}C_4 + ^{47}C_3) + ^{4}C_3 +^{49}C_3 +^{50}C_3 +^{51}C_3 $$
$$ = (^{48}C_4 +^{48}C_3 + ^{49}C_3 +^{50}C_3 +^{51}C_3 $$
$$( \because ^nC_r + ^nC_{r-1} = ^{n+1}C_r) $$
$$ = (^{49}C_4 +^{49}C_3) + ^{50}C_3 + ^{51}C_3)( \because ^nC_r + ^nC_{r-1} = ^{n+1}C_r) $$
$$ =(^{50}C_4 + ^{50}C_3) + ^{51}C_3 )(\because ^nC_r +^nC_{r-1} = ^{n+1}C_r) $$
$$ =(^{51}C_4 + ^{51}C_3 \quad ( \because ^nC_r +^nC_{r-1} = ^{n+1}C_r )$$
$$ =^{52}C_4 \quad ( \because ^nC_r + ^nC_{r-1} = ^{n+1}C_r )$$
If $$_{ }^{ n+1 }{ { C }_{ r+1 } }$$: $$_{ }^{ n }{ { C }_{ r } }$$: $$_{ }^{ n-1 }{ { C }_{ r-1 } }=$$ 11: 6: 3 , then $$r=$$
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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 $$n=10$$
lf $$\displaystyle x,y\in(0,30)$$ 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
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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 $$=4\times7=28$$
If $$ n = ^mC_2 $$, then the value of $$ ^nC_2 $$ is given by
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0%
$$ ^{m+1}C_4 $$
0%
$$ ^{m-1}C_4 $$
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$$ ^{m+2}C_4 $$
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$$ 3.^{m+1}C_4 $$
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 $$\mathrm{S}_{2}=10\times 2^{8}$$.
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Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1
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Statement-1 is true, Statement-2 is false
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Statement-1 is false, Statement-2 is true
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Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1
Explanation
$$
\mathrm{S}_{1}=\sum_{\mathrm{j}=1}^{10}|(j-1)\dfrac{10!}{(j-1)(j-2)!(10-j)!}=90\sum_{\mathrm{j}=2}^{10}\dfrac{8!}{(j-2)!(8-(j-2))!}=90\cdot 2^{8}
$$
$$
\mathrm{S}_{2}=\sum_{\mathrm{j}=1}^{10}|\dfrac{10!}{j(j-1)!(9-(j-1))!}=10\sum_{\mathrm{j}=1}^{10}\dfrac{9!}{(j-1)!(9-(j-1))!}=10\cdot 2^{9}
$$
$$\displaystyle \mathrm{S}_{3}=\sum_{\mathrm{j}=1}^{10}[j(j-1) +j]\displaystyle \dfrac{10!}{|!(10-j)!}=\sum_{\mathrm{j}=1}^{10}|(j-1)^{10}\mathrm{C}_{\mathrm{j}}=\sum_{\mathrm{j}=1}^{10}j^{10}\mathrm{C}_{\mathrm{j}}=90.2^{8}+10.2^{9}$$
$$=90.2^{8}+20.2^{8}=110.2^{8}=55.2^{9}$$.
If $$ ^{n-1}C_r=(k^2-3 ) ^nC_{r+1} $$, then $$ k\in $$
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0%
$$ (\infty ,-2) $$
0%
$$ (2,\infty ) $$
0%
$$ [-\sqrt { 3 } ,\sqrt { 3 } ] $$
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$$ (\sqrt { 3 } ,2] $$
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
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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 $$\dfrac{1.3.5....(2n-1)}{2.4.6....(2n)}$$?
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0%
$$(2n!) \div (2^n(n!))^2$$
0%
$$(2n!) \div n!$$
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$$(2n - 1) \div (n - 1)!$$
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$$2^n$$
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
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0%
$$x = 4$$
0%
$$x = 5$$
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$$x= 6$$
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$$x= 7$$
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
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$$n+1$$
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$$n/2$$
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$$n+2$$
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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?
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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?
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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.
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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
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TOJECB
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TOEJBC
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TOCJEB
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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 $$^{47}C_4 + \displaystyle\sum_{r=1}^{5}{^{52-r}C_3}$$=
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$$^{52}C_2$$
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$$^{52}C_3$$
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$$^{52}C_4$$
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$$^{52}C_5$$
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
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$$p$$ divides $$N$$ completely
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$$p^2$$ divides $$N$$ completely
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$$p$$ cannot divide $$N$$
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none of these
Explanation
Solution:
$$N={}^{2n}C_n=\cfrac{2n!}{n!n!}=\cfrac{(n+1)(n+2)(n+3)........(2n)}{(1).(2).......(n)}$$
As $$p>n$$ 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
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$$468$$
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
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$$8$$
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 $$n=7, 8, .....$$
$$^nC_r + \displaystyle\sum_{j=0}^{3}{^{n+j}C_{r+1+j}} = $$_________________
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$$^{n+3}C_{r+3}$$
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$$^{n}C_{r+4}$$
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$$^{n+1}C_{r+4}$$
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$$^{n+4}C_{r+4}$$
Explanation
Expanding
We get
$$\:^{n}C_{r}+\:^{n}C_{r+1}+\:^{n+1}C_{r+2}+\:^{n+2}C_{r+3}+\:^{n+3}C_{r+4}$$
$$=\:^{n+1}C_{r+1}+\:^{n+1}C_{r+2}+\:^{n+2}C_{r+3}+\:^{n+3}C_{r+4}$$
$$=\:^{n+2}C_{r+2}+\:^{n+2}C_{r+3}+\:^{n+3}C_{r+4}$$
$$=\:^{n+3}C_{r+3}+\:^{n+3}C_{r+4}$$
$$=\:^{n+4}C_{r+4}$$
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Practice Class 11 Engineering Maths Quiz Questions and Answers
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