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CBSE Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 13 - MCQExams.com
CBSE
Class 11 Engineering Maths
Permutations And Combinations
Quiz 13
The value of $${^{50}C_4}+\sum^6_{r=1}{^{56-r}C_3}$$ is?
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0%
$$^{55}C_4$$
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$$^{55}C_3$$
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$$^{56}C_3$$
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$$^{56}C_4$$
If $$^{32}C_{2n-1}=^{32}C_{n-3}$$ then $$n=$$
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0%
$$10$$
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$$9$$
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$$12$$
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$$11$$
The value of $${ C }_{ 0 }^{ 2 }+3{ C }_{ 1 }^{ 2 }+5{ C }_{ 2 }^{ 2 }+.....$$ to $$n+1$$ terms is
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$$n{ \cdot ^{2n}}{C_n}$$
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$$(2n+2)\cdot ^{ 2n }{ C }_{ n }$$
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$$(n + 1){ \cdot ^{2n}}{C_{n - 1}}$$
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$$(2n + 1){ \cdot ^{2n - 1}}{C_{n - 1}}$$
The number of ways is which an examiner can assign $$30$$ marks to $$8$$ questions, giving not less
Then $$2$$ marks to any question is-
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$$^{21}C_{7}$$
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$$^{21}C_{8}$$
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$$^{21}C_{9}$$
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$$^{21}C_{10}$$
If $$(1+x)^{15}=C_{0}+C_{1}xC_{2}x^{2}+...+C_{15}x^{15},$$ then $$^{15}C_{0}^{2}- ^{15}C_{1}^{2}+^{15}C_{2}^{2}- ^{15}C_{2}^{3}+... ^{15}C_{15}^{2}$$ is equal to
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0%
0
0%
1
0%
-1
0%
none of these
$$\text{The number of ways dividing 52 cards amongst four players equally, are} $$
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$$\dfrac {52!}{(13!)^{4}}$$
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$$\dfrac {52!}{(13!)^{4}4!}$$
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$$\dfrac {52!}{(12!)^{4}4!}$$
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$$\text{None of these}$$
Explanation
$$\textbf{Hint:-Number of ways in which 3n diiferent things can be divided equally three people }$$
$$=\large\bf\frac{(3n)!}{n!n!n!}$$
$$\textbf{Step:-1 Find the total number of ways }$$
$$\text{Here 52 cards are distributed among four players equally in an order so each gets 13 cards}$$
$$\text{Total number of ways dividing 52 cards amongst four players equally,are=}$$ $$^{52}C_{13}\ \times \ ^{39}C_{13}\ \times$$$$\ ^{26}C_{13}\ \times \ ^{13}C_{13}$$
$$=\large\frac{52!}{39!13!}$$$$\times\large\frac{39!}{26!13!}$$$$\times\large\frac{26!}{13!13!}$$$$\times\large\frac{13!}{0!13!}$$
$$=\large\frac{52!}{13!\times13!\times13!\times13!}$$
$$=\large\frac{52!}{{13!}^4}$$
$$\textbf{Heence the number of ways the cards can be divided equally amongst number of players=}$$$$\large\bf\frac{52!}{{13!}^4}$$
If $$^{n}C_{1}=^{n}C_{2}$$, then $$n$$ is equal to
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0%
$$2$$
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$$3$$
0%
$$5$$
0%
$$none\ of\ these$$
$${ C }_{ 0 }+({ C }_{ 0 }+{ C }_{ 1 })+({ C }_{ 0 }+{ C }_{ 1 }+{ C }_{ 2 })+......+({ C }_{ 0 }+{ C }_{ 1 }+........+{ C }_{ n })=$$
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$$n.{ 2 }^{ n-1 }$$
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$$(n+2).{ 2 }^{ n }$$
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$${ 2 }^{ n }$$
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$$(n+2).{ 2 }^{ n-1 }$$
If $$^{n}C_{r}=\dfrac {n\ !}{r\ !(n-r)\ !}$$ then the sum of the series $$1+^{n}C_{1}+^{n+1}C_{2}+^{n+2}C_{3}+.....+^{n+r-1}C_{r}$$ is equal to
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$$^{n+r}C_{r}$$
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$$^{n+r-1}C_{r}$$
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$$^{n+r}C_{r-1}$$
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$$None\ of\ these$$
If the second term of the expansion $$\left[a^{1/13}+\dfrac{a}{\sqrt{a^{-1}}}\right]^n$$ is $$14a^{5/2}$$ then the value of $$\dfrac{^{n}C_3}{^{n}C_2}$$ is
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0%
$$4$$
0%
$$3$$
0%
$$12$$
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$$6$$
Explanation
2nd Term $$= ^nC_1(a^{1/13})^{(n-1)} \left ( \frac{a}{\sqrt{a-1}} \right )^1=14 a^{5/2}$$
$$\Rightarrow n\,\,\, a^{\tfrac{n-1}{13}}.a^{3/2}= 14a^{5/2}$$
$$\Rightarrow $$ Comparing both sides
$$\Rightarrow \dfrac{n-1}{13}+\dfrac{3}{2}=\dfrac{5}{2}$$
$$\Rightarrow \dfrac{n-1}{13}=1\Rightarrow n-1=13$$
$$\Rightarrow n=14$$
$$\dfrac{^nC_3}{^nC_2}\Rightarrow \dfrac{\dfrac{n!}{3!(n-3)!}}{\dfrac{n!}{2!(n-2)!}}=\dfrac {2!(n-2)!n!} {n!3!(n-3)!}$$
$$\Rightarrow \dfrac {2!12!}{3! 11!}=\dfrac {12}{3}=4$$
If $$\dfrac{c_0}{2}-\dfrac{c_1}{3}+\dfrac{c_2}{4}-.....+(-1)^n\dfrac{c_n}{n+2}=\dfrac{1}{2013\times2014}$$ then n =
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$$2013$$
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$$2014$$
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$$2012$$
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$$2011$$
If $$C_{r}$$ stands for $$^{n}C_{r}$$ then the sum of the series $$\frac{2(\frac{n}{2})!(\frac{n}{2})!}{n!}[C_{0}^{2}-2C_{1}^{2}+3C_{2}^{2}-.....+(-1)^{n}(n+1)C_{n}^{2}]$$,
where n is an even positive integer, is
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$$(-1)^{n/2}(n+2)$$
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$$(-1)^{n}(n+2)$$
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$$(-1)^{n/2}(n+1)$$
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None of these
$$\sum _{ r=0 }^{ n }{ r\left( n-r \right) \left( _{ }^{ n }{ { C }_{ r } } \right) ^{ 2 } } $$ is equal to
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$$n^{2}._{ }^{ 2n-1 }{ C }_{ n } $$
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$$n^{2}._{ }^{ 2n }{ { C }_{ n-1 } }$$
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$$_{ }^{ 2n-1 }{ { C }_{ n-1 } }$$
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None of these
How many different $$5$$ letter sequences can be made using the letters $$A,B,C,D$$ with repetition such that the sequence does not include the word $$BAD$$?
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$$1024$$
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$$976$$
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$$48$$
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$$678$$
If $$^{50}C_{r}=C_{r}$$ then the value of $$C_{0}^{2}+\frac{1}{2}C_{1}^{2}+\frac{1}{3}C_{2}^{2}+\frac{1}{4}C_{3}^{2+....+\frac{1}{50}^{101}C_{51}}$$
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$$\frac{1}{51}^{101}C_{50}$$
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$$\frac{1}{50}^{101}C_{50}$$
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$$\frac{1}{51}^{101}C_{49}$$
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$$\frac{1}{50}^{101}C_{51}$$
$$^n$$C$$_0$$ + 2$$^2$$ . $$\dfrac{^nC_1}{2}$$ + 2$$^3$$ . $$\dfrac{^nC_2}{3}$$ +.......+ 2$${n + 1}$$ . $$\dfrac{^nC_n}{n + 1}$$ =
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3$$^{n+ 1} $$ - 1
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$$\dfrac{3^{n+1} - 1}{n + 1}$$
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$$\dfrac{3^{n +1}}{n + 1}$$
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$$\dfrac{3^{n +1 }-1}{n}$$
The number of ways in which three girls and ten boys can be seated in two vans, each having numbered seats, three in the front and four at the back is
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$${ ^{ 14 }C }_{ 12 }$$
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$${ ^{ 14 }P }_{ 13 }$$
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$$14!$$
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$${ 10 }^{ 3 }$$
The sum $$\sum _{ r=1 }^{ n }{ r.{ }^{ 2n }{ C }_{ r } } $$ is equal to.
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$$n.2^{2n-1}$$
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$$2^{2n-1}$$
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$$2^{n-1}+1$$
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none of these
The total no. of six digit numbers $${ x }_{ 1 }{ x }_{ 2 }{ x }_{ 3 }{ x }_{ 4 }{ x }_{ 5 }{ x }_{ 6 }$$ having the property $${ x }_{ 1 }<{ x }_{ 2 }\le { x }_{ 3 }<{ x }_{ 4 }<{ x }_{ 5 }\le { x }_{ 6 }$$, is equal to
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$$_{ }^{ 11 }{ C }_{ 6 }$$
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$$_{ }^{ 16 }{ C }_{ 2 }$$
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$$_{ }^{ 17 }{ C }_{ 2 }$$
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$$_{ }^{ 18 }{ C }_{ 2 }$$
If $$^nC_r$$ denotes the number of combinations of n things taken r things at a time, then the expression $$^nC_{r+1} + ^nC_{r-1}+2 ^nC_r is$$
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$$^{n+2}C_{r+1}$$
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$$^{n+1}C_r$$
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$$^{n+1}C_{r+1}$$
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$$^{n+2}C_r$$
$$\dfrac { { C }_{ 1 } }{ { C }_{ 0 } } +\dfrac { 2{ C }_{ 2 } }{ { C }_{ 1 } } +...\dfrac { 3{ C }_{ 3 } }{ { C }_{ n-1 } } =$$
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$$\dfrac { n\left( n-1 \right) }{ 2 } $$
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$$\dfrac { n\left( n+2\right) }{ 2 } $$
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$$\dfrac { n\left( n+1 \right) }{ 2 } $$
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$$\dfrac { \left( n-1 \right) \left( n-2 \right) }{ 2 } $$
If $$\sum\limits_{r = 0}^n {\left( {\frac{{r + 2}}{{r + 1}}} \right)} \,{\,^n}{C_r} = \frac{{{2^8} - 1}}{6},$$ then 'n' is
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0%
8
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4
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6
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5
The value $$\sum _{ r=0 }^{ 10 }{ r_{ }^{ 10 }{ C_{ r } } } .{ \left( -2 \right) }^{ 10-r }$$ is
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0%
10
0%
20
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30
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300
$$\dfrac { { C }_{ 1 } }{ { C }_{ 0 } } +\dfrac { 2{ C }_{ 2 } }{ { C }_{ 1 } } +\dfrac { { 3C }_{ 3 } }{ { C }_{ 2 } } +.....+\dfrac { { nC }_{ n } }{ { C }_{ n-1 } } =$$
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0%
$$\dfrac { n(n-1) }{ 2 } $$
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$$\dfrac { n(n+2) }{ 2 } $$
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$$\dfrac { n(n+1) }{ 2 } $$
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$$\dfrac { (n-1)(n-2) }{ 2 } $$
If $$(1+x)^{n}=C_{0}+C_{1}x+C_{2}x^{2}+...+C_{n}x^{n}$$, then the value of $$C_{0}^{2}+\dfrac{C_{1}^{2}}{2}+\dfrac{C_{2}^{2}}{2}+.........+\dfrac{C_{n}^{2}}{n+1}$$ is
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0%
$$\dfrac{(2n-1)!}{\left\{(n+1)!\right\}^{2}}$$
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$$\dfrac{(2n-1)!}{(n+1)!}$$
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$$\dfrac{(2n+1)!}{\left\{(n+1)!\right\}^{2}}$$
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$$\dfrac{(2n)!}{(n+1)!}$$
The number of odd numbers lying between 40000 and 70000 that can be made from the digits 0, 1, 2, 4, 5, 7 if digits can be repeated any number of times is
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1125
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1296
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766
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655
If $${a_n} = \sum\limits_{r = 0}^n {\frac{1}{{^n{C_r}}},} $$ then $${a_n} = \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}},} $$ equals
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0%
$$^{\left( {n - 1} \right)}{a_n}$$
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$$^n{a_n}$$
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$$\frac{n}{2}{a_n}$$
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None of these
$${ }^{ n-2 }{ C }_{ r }+2{ }^{ n-2 }{ C }_{ r-2 }+{ }^{ n-2 }{ C }_{ r-2 }$$ equals:
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0%
$${ }^{ n+1 }{ C }_{ r }$$
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$${ }^{ n }{ C }_{ r }$$
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$${ }^{ n }{ C }_{ r+1 }$$
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$${ }^{ n-1 }{ C }_{ r }$$
If $$^{ n }{ C }_{ 3 }=^{ n }{ C }_{ 2 }$$, then n is equal to
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0%
2
0%
3
0%
5
0%
none of these
$$If\,a\, = {\,^m}{C_2},\,\,then{\,^a}{C_{2\,}}\,is\,equal\,to$$
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0%
$$^{m + 1}{C_4}$$
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$$^{m - 1}{C_4}$$
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$$3{ \cdot ^{m + 2}}{C_4}$$
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$$3{ \cdot ^{m + 1}}{C_4}$$
$$\sum _{ r=0 }^{ n-1 }{ \frac { { }^{ n }{ C }_{ r } }{ { }^{ n }{ C }_{ r }+{ }^{ n }{ C }_{ r+1 } } } $$ is equal to :
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$$\frac n2$$
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$$\frac {n+1}2$$
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$$\frac {n(n+1)}2$$
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$$\frac {n(n-1)}{2n(n+1)}$$
If $$^{2n}C_3$$ : $$^{n}C_2$$ : : $$44 : 1$$, then the value of n is
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0%
17
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6
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11
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none of these
Total number of 6-digit numbers in which all the odd digits and only odd digits appears, is
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$$\frac{5}{2}\left( {6!} \right)$$
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$${6!}$$
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$$\frac{1}{2}\left( {6!} \right)$$
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$$\frac{3}{2}\left( {6!} \right)$$
$$\sum\limits_{r = 1}^n {\frac{{r{.^n}{C_r}}}{{^n{C_{r - 1}}}}} $$ is equal to:
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$$\frac{{n\left( {n + 1} \right)}}{2}$$
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$$\frac{{n\left( {n - 1} \right)}}{2}$$
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$${n\left( {n + 1} \right)}$$
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$${n\left( {n - 1} \right)}$$
The value of the expression $${ }^{ n+1 }{ C }_{ 2 }+2[{ }^{ 2 }{ C }_{ 2 }+{ }^{ 3 }{ C }_{ 2 }+{ }^{ 4 }{ C }_{ 2 }+.....+{ }^{ n }{ C }_{ 2 }]$$ is
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0%
$$\sum { n } $$
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$$\sum { n^2 }$$
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$$\sum { n^3 } $$
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$$\frac {(n+1)}{2}$$
The value of $${ }^{ 15 }{ C }_{ 8 }+{ }^{ 15 }{ C }_{ 9 }-{ }^{ 15 }{ C }_{ 6 }-{ }^{ 15 }{ C }_{ 7 }$$ is :
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-1
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0
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1
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None of these
The value of $$\frac{{^{11}{C_0}}}{1} + \frac{{^{11}{C_1}}}{2} + \frac{{^{11}{C_2}}}{3} + ....... + \frac{{^{11}{C_{11}}}}{{12}}$$ will be
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$$\frac{1}{{12}}\left( {{2^{12}} - 1} \right)$$
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$$\frac{1}{{11}}\left( {{2^{12}} - 1} \right)$$
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$$\frac{1}{{12}}\left( {{2^{12}} + 1} \right)$$
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none of these
How many combinations can be formed of 8 counters marked 1 2 ..... 8 taking 4 at a time there being at lest one odd and even numbered in each combination?
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68
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66
0%
64
0%
62
The value of $$^{50}{C_4} + \sum\limits_{r = 1}^6 {^{56 - r}{c_3}} $$ is
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0%
$$^{55}{c_4}$$
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$$^{55}{c_3}$$
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$$^{56}{c_3}$$
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$$^{56}{c_4}$$
Which of the following is not true?
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$${ C } _ {{ r } } = ^ { n } C _ { { n } - { r } }$$
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$$^ { n } C _ { r } + ^ { n } C _ { r - 1 } = ^ { n + 1 } C _ { r }$$
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$$r . ^ { n } C _ { r } = n . ^ { n - 1 } C _ { r - 1 }$$
0%
$$ { C } _ { { r } } + ^ { { n } } { C } _ { { r } - 1 } = ^ { { n } } { C } _ { { r+1 } }$$
$$\sum _{ r=1 }^{ n }{ \left\{ \sum _{ { r }_{ 1 }=0 }^{ r-1 }{ ^{ n }{ C }_{ r }^{ n }{ C }_{ { r }_{ 1 } }{ 2 }^{ { r }_{ 1 } } } \right\} } =$$
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0%
$${4}^{n}-{3}^{n}+1$$
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$${4}^{n}-{3}^{n}-1$$
0%
$${4}^{n}-{3}^{n}+2$$
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$${4}^{n}-{3}^{n}$$
$$
^{47} C_{r}+\sum_{j=1}^{5}(52-1) C_{3}=
$$
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0%
$$
^{52} \mathrm{C}_{4}
$$
0%
$$
^{52} \mathrm{C}_{3}
$$
0%
$$
^{53} \mathrm{C}_{4}
$$
0%
$$
^{53} \mathrm{C}_{3}
$$
If $$_{ }^{ n-1 }{ { C }_{ r } }=({ k }^{ 2 }-3)_{ }^{ n }{ { C }_{ r+1 } }$$, then k belongs to :
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0%
$$(-\infty ,-2]$$
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$$[2,\infty )$$
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$$[-\sqrt { 3 } ,\sqrt { 3 } ]$$
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$$(\sqrt { 3 } ,2]$$
If $$\sum\limits_{r = m}^n {^r{C_m}{ = ^{n + 1}}{C_{m + 1,}}} $$ then $$\sum\limits_{r = m}^n {{{\left( {n - r + 1} \right)}^r}{C_m}} $$ is equal to
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0%
$$^{n + 3}{C_{m + 2}}$$
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$$^{n + 2}{C_{m + 2}}$$
0%
$$^{n + 2}{C_{m + 1}}$$
0%
none of there
$$\left( ^ { 2 n } C _ { 0 } \right) ^ { 2 } - \left( ^ { 2 n } C _ { 1 } \right) ^ { 2 } + \left( ^ { 2 n } C _ { 2 } \right) ^ { 2 } - \ldots + ( - 1 ) ^ { 2 n } \left( ^ { 2 n } C _ { 2 n } \right) ^ { 2 }$$ equals to
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0%
$$\left( ^ { 2 n } C _ { n } \right) ^ { 2 }$$
0%
$$( - 1 ) ^ { 2 n } \left( ^ { 2 n } C _ { n } \right)$$
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$$( - 1 ) ^ { n } \left( ^ { 2 n } C _ { n } \right)$$
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$$\left( ^ { n } C _ { n } \right) + \left( ^ { 2 n } C _ { n } \right)$$
If $${C_r} = {\,^n}{C_r}\,\,and\,\,\left( {{C_0} + {C_1}} \right)\,\,\left( {{C_1} + {C_2}} \right)......$$
$$\left( {{C_{n - 1}} + {C_n}} \right) = k\frac{{{{\left( {n + 1} \right)}^n}}}{{n!}}$$ then the value of k is
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0%
$${C_0}.{C_1}.{C_2}....{C_n}$$
0%
$${\left( {{C_1}} \right)^2}.{\left( {{C_2}} \right)^2}....{\left( {{C_n}} \right)^2}$$
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$${C_1} + {C_2} + .... + {C_n}$$
0%
None of these
Total number of four-digit odd numbers that can be formed using 0, 1, 2, 3, 5, 7 (using repetition allowed) are
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0%
216
0%
375
0%
400
0%
720
In a conference hall 10 executive are to be seated on 10 chairs placed left to right. Executive $$E_1$$ wants to sit to the right of $$E_2$$ Also, $$E_4$$ wants to sit the left of $$E_1$$ and $$E_5$$ wants to sits to the left of ways $$E_4$$ Number of ways the executives can be seated on the 10 chairs, is
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0%
$$\dfrac{10!}{4!}$$
0%
$$\dfrac{10!}{3}$$
0%
$$10!$$
0%
$$3 \times \dfrac{10!}{4!}$$
Number of positive integers $$n ,$$ less than $$17 ,$$ for which $$n ! + ( n + 1 ) ! + ( n + 2 ) !$$ is an integral multiple of $$49$$ is
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0%
$$2$$
0%
$$5$$
0%
$$7$$
0%
$$9$$
If $$^{35}C_{n+7}=^{35}C_{4n-2}$$, then the value of $$n$$ is
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0%
$$3$$
0%
$$4$$
0%
$$5$$
0%
$$6$$
Explanation
Given
$$^{35}C_{n+7}=^{35}C_{4n-2}$$
It is in form if
$$^{n}C_{x}=^{n}C_{v}$$, then
x=y or x+y=n
now
(i)
n+7=4n-2
4n-n=7+2
3n=9
n=3
(ii)
n+7+4n-2=35
5n+5=35
5n=30
n=6
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