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CBSE Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 14 - MCQExams.com
CBSE
Class 11 Engineering Maths
Permutations And Combinations
Quiz 14
What is the hundreds digits of ( 20 !-15!) ?
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1
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4
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5
$$\left( \begin{matrix} 51 \\ 3 \end{matrix} \right) +\left( \begin{matrix} 50 \\ 3 \end{matrix} \right) +\left( \begin{matrix} 49 \\ 3 \end{matrix} \right) +\left( \begin{matrix} 48 \\ 3 \end{matrix} \right) +\left( \begin{matrix} 47 \\ 3 \end{matrix} \right) +\left( \begin{matrix} 47 \\ 4 \end{matrix} \right) $$ is equal to (where $$\left( \begin{matrix} n \\ r \end{matrix} \right) $$ denotes $$_{ }^{ n }{ { C }_{ r } }$$
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$$\left( \begin{matrix} 52 \\ 1 \end{matrix} \right) $$
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$$\left( \begin{matrix} 52 \\ 2 \end{matrix} \right) $$
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$$\left( \begin{matrix} 52 \\ 3 \end{matrix} \right) $$
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$$\left( \begin{matrix} 52 \\ 4 \end{matrix} \right) $$
If $$n$$ and $$r$$ are positive integers such that $$r < n$$ then $$^{n}C_{r}+^{n}C_{r-1}$$ is equal to
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$$^{2n}C_{2r-1}$$
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$$^{n+1}C_{r}$$
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$$^{n}C_{r+1}$$
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$$2^{n}C_{r}$$
If $$x\epsilon N and C^{x-1_{4}}-C^{X-1}_{3}-\frac{5}{4}P^{x-2}_{2}< 0$$ then x is equal to
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4,5,6,7,8,9,10
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3
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$$3\leq x\leq 10$$
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none of these
If $$m$$ and $$n$$ are positive integers more than or equal to $$2 , m > n ,$$ then $$( m n ) !$$ is divisible by
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$$( m ! ) ^ { n }$$
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$$( n ! ) ^ { m }$$
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$$( m + n ) !$$
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$$( m - n ) !$$
If $${ n }_{ C_ r }$$ denotes the number of combinations of n things taken r at a time, then the expression + $${n}_{C_r-1} + 2 \times {n}_{C_r}$$ equals
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$${ n+2 }_{ C_ r }$$
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$${ n+2 }_{ C_ r+1 }$$
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$${ n+2 }_{ C_ r }$$
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$${ n+2 }_{ C_ r+1 }$$
If $$^{189}C_{35} + ^{189}C_{x} = ^{190}C_{x}$$, then $$x$$ is equal to
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$$34$$
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$$35$$
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$$36$$
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$$37$$
In a village, there are $$87$$ families of which $$52$$ families have at most $$2$$ children. In a rural development programme, $$20$$ families are to be helped chosen for assistance, of which at least $$18$$ families must have at most $$2$$ children. In how many ways can the choice be made?
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$$\ ^{52}C_{18}\times \ ^{35}C_{2}+\ ^{52}C_{19}\times \ ^{35}C_{1}+\ ^{52}C_{20}\times \ ^{35}C_{0}$$
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$$\ ^{52}C_{18}\times \ ^{35}C_{2}+\ ^{52}C_{19}\times \ ^{35}C_{1}+\ ^{52}C_{20}\times \ ^{35}C_{10}$$
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$$\ ^{52}C_{18}\times \ ^{35}C_{2}+\ ^{52}C_{19}\times \ ^{35}C_{1}+\ ^{52}C_{2}\times \ ^{5}C_{0}$$
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$$\ ^{52}C_{18}\times \ ^{35}C_{2}+\ ^{52}C_{9}\times \ ^{35}C_{10}+\ ^{52}C_{20}\times \ ^{5}C_{0}$$
A sports team of $$11$$ students is to be constituted, choosing at least $$5$$ from class $$XI$$ and at least $$5$$ from class $$XII$$. If there are $$20$$ students in each of these classes, in how many ways can the teams be constituted?
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$$2(\ ^{20}C_{3}\times \ ^{20}C_{7})$$
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$$2(\ ^{20}C_{5}\times \ ^{20}C_{6})$$
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$$2(\ ^{20}C_{4}\times \ ^{20}C_{8})$$
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None of these
Explanation
A team of $$11$$ students can be constituted in the following two ways
$$i) 5$$ student from class $$XI$$ and $$6$$ from $$XII$$
$$ii) 6$$ student from class $$XI$$ and $$5$$ from $$XII$$
$$\therefore$$ The required number of ways
$$\Rightarrow \ ^{20}C_{5}\times \ ^{20}C_{6}+\ ^{20}C_{6}\times \ ^{20}C_{5}$$
$$\Rightarrow 2\left(\ ^{20}C_{5}\times \ ^{20}C_{6}\right)$$
The number of five digit numbers in which digits decrease from left to right, is
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$$^{9}C_{5}$$
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$$^{10}C_{5}$$
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$$2\times ^{10}C_{5}$$
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$$^{10}P_{5}$$
If $$\dfrac {^{n}C_{r} + 4^{n}C_{r + 1} + 6^{n}C_{r + 2} + 4^{n}C_{r + 3} + ^{n}C_{r + 4}}{^{n}C_{r} + 3^{n}C_{r + 1} + 3^{n}C_{r + 2} + ^{n}C_{r + 3}} = \dfrac {n + k}{r + k}$$ then the value of $$k$$ is
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$$1$$
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$$2$$
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$$4$$
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$$5$$
The number of four -digit numbers strictly grater than $$4321$$ that can be formed using the digits $$0,1,2,3,4,5$$ (repetition of digits is allowed) is:
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$$288$$
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$$306$$
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$$360$$
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$$310$$
Explanation
(1) The number of four-digit numbers Strating with 5 is equal to $$6^3 = 216$$
(2) Starting with 44 and 55 is equal to $$36\times 2 = 72$$
(3) Starting with 433, 434 and 435 is equal to $$6\times 3 = 18$$
(4) Remaining numbers are 4322, 4323, 4324, 4325 is equal to 4
so total number are
$$716 + 72 + 18 + 4 = 310$$
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