Permutations and Combinations Test 12

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Permutations and Combinations Test 12

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Weekly Quiz Competition

Question 1
1 / -0

$$\displaystyle \begin{pmatrix} 47 \\ 4 \end{pmatrix}$$+$$\displaystyle \sum_{j=1}^{5}$$ $$\displaystyle \begin{pmatrix} 52-j \\ 3 \end{pmatrix}$$=$$\displaystyle \begin{pmatrix} x \\ y \end{pmatrix}$$ then $$\displaystyle \frac{x}{y}$$ =

A
11

B
12

C
13

D
14

Solution

Expanding, we get
$$\:^{47}C_{4}+\:^{47}C_{3}+\:^{48}C_{3}+\:^{49}C_{3}+\:^{50}C_{3}+\:^{51}C_{3}+\:^{52}C_{3}$$
Applying
$$\:^{n}C_{r}+\:^{n}C_{r+1}=\:^{n+1}C_{r+1}$$, we get
$$[\:^{47}C_{4}+\:^{47}C_{3}+\:^{48}C_{3}+\:^{49}C_{3}+\:^{50}C_{3}+\:^{51}C_{3}]$$
$$=+\:^{48}C_{4}+\:^{48}C_{3}+\:^{49}C_{3}+\:^{50}C_{3}+\:^{51}C_{3}]$$
:
:
$$\:^{51}C_{3}+\:^{51}C_{4}$$
$$=\:^{52}C_{4}$$.
Hence $$x=52$$ and $$y=4$$
Hence, $$\dfrac {x}{y}=\dfrac{52}{4}$$
$$=13$$
Question 2
1 / -0

The mean of the values $$0, 1, 2, \cdots n$$, having corresponding weights $$^nC_0, ^nC_1, \cdots ^nC_n,$$ respectively is

A
$$\cfrac{n+1}{2}$$

B
$$\cfrac{2^{n-1}+1}{n+1}$$

C
$$\cfrac{2^n}{n}$$

D
$$\cfrac{n}{2}$$

Solution

Required A.M $$=\cfrac{\sum x_if_i}{\sum f_i} = \cfrac{0.^nC_0+^nC_1+2.

^nC_2+3. ^nC_3+..............+n. ^nC_n}{^nC_0+^nC_1+ ^nC_2+

^nC_3+.......+ ^nC_n} = \mu $$ (say)
Now consider $$(1+x)^n =^nC_0+^nC_1 x+ ^nC_2 x^2+.........+^nC_n x^n ....(1)$$
Differentiating both side (1) w.r.t $$x$$
$$n(1+x)^{n-1} =^nC_1 +2. ^nC_2 x+.........+n. ^nC_n x^{n-1} ....(2)$$
Now putting $$x=1$$ in (1) and (2) we get,
$$^nC_0+^nC_1 + ^nC_2 +.........+^nC_n =2^n$$
and $$ ^nC_1 +2. ^nC_2 +.........+n. ^nC_n = n. 2^{n-1}$$
$$\therefore \mu = \cfrac{n. 2^{n-1}}{2^n}=\cfrac{n}{2}$$
Question 3
1 / -0

An $$n-$$ digit number is a positive number with exactly $$n$$ digits. Nine hundred distinct n−digit numbers are to be formed using only the three digits 2,5 and 7. The smallest value of $$n$$ for which this is possible is

A
$$5$$

B
$$6$$

C
$$7$$

D
$$8$$

Solution

Using $$2,5$$ and $$7$$ with repetition each place of $$n$$ digit number can be chosen in $$3$$ ways.
Hence, total number of $$n-$$digit numbers $$=3\times 3\times 3...n$$ times $$={3}^{n}.$$
According to given condition $${ 3 }^{ n }\ge 900\Rightarrow { 3 }^{ n-2 }\ge 100$$
$$\therefore n-2\ge 5\Rightarrow n\ge 7$$
Question 4
1 / -0

If $$^{20}C_{r\, +\, 2}\, =\, ^{20}C_{2r\, -\, 3}\,$$, find $$\, ^{12}C_r$$.

A
$$792$$

B
$$795$$

C
$$790$$

D
None of these

Solution

$$^{20}C_{r+2} = ^{20}C_{2r-3}$$

$$\Rightarrow r+2 = 2r-3$$, or$$ r+2=20-2r+3$$ [ Using $$^nC_x=^nC_y=> x=y\ or x=n-y$$ ]

$$\Rightarrow r = 5 \ or \ r=7 \therefore ^{12}C_5 = ^{12}C_7=792$$
Question 5
1 / -0

Directions For Questions

S = {0, 2, 4, 6, 8}. A natural number is said to be divisible by 2 if the digit at the unit place is an even number.The number is divisible by 5, if the number at the unit place is 0 or 5. If four numbers are selected from S and a four digit number ABCD is formed.
On the basis of above information, answer the following questions

...view full instructions

The number of such numbers which are divisible by two and five (all digits are not different) is

A
125

B
76

C
65

D
100

Solution

The numbers which are divisible by 2 and 5 are the multiples of 10.
Hence the ending digit has to be 0.
However the starting digit cannot be 0.
Now
_ _ _ 0.
Hence number of ways of filling the first place will be 4, the second place will be 5 and 3rd place will be 5, since repetition of digits are allowed.
Hence total number of the numbers which are divisible by 2 and 5 are
$$4\times 5\times 5$$
$$=100$$
Question 6
1 / -0

A set contains $$(2n + 1)$$ elements. The number of subsets of the set which contain at most n elements is

A
$$2^{n}$$

B
$$2^{n\, +\, 1}$$

C
$$2^{n\, -\, 1}$$

D
$$2^{2n}$$

Solution

Let $$N$$ = the number of subsets of the set which contain at most n elements

$$=^{ 2n+1 }{ { C }_{ 0 } }+^{ 2n+1 }{ { C }_{ 1 } }+^{ 2n+1 }{ { C }_{ 2 } }+...+^{ 2n+1 }{ { C }_{ n } }$$

$$ 2N=2\left( ^{ 2n+1 }{ { C }_{ 0 }+^{ 2n+1 }{ { C }_{ 1 }+^{ 2n+1 }{ C_{ 2 }+...+^{ 2n+1 }{ { C }_{ n } } } } } \right) $$

$$ =\left( ^{ 2n+1 }{ { C }_{ 0 }+^{ 2n+1 }{ { C }_{ 2n+1 } } } \right) +\left( ^{ 2n+1 }{ { C }_{ 1 }+ }^{ 2n+1 }{ { C }_{ 2n } } \right) +...+\left( ^{ 2n+`1 }{ { C }_{ n }+^{ 2n+1 }{ { C }_{ n+1 } } } \right) $$........................$$ \left( \because ^{ n }{ { C }_{ r }= }^{ n }{ C }_{ n-r } \right) $$

$$ =^{ 2n+1 }{ C }_{ 0 }+^{ 2n+1 }{ C }_{ 1 }+^{ 2n+1 }{ { C }_{ 2 } }+...+^{ 2n+1 }{ { C }_{ 2n+1 } }$$

$$ { =2 }^{ 2n+1 }$$

$$ \Rightarrow 2N={ 2 }^{ 2n+1 }$$

$$ \Rightarrow N={ 2 }^{ 2n }$$
Question 7
1 / -0

If $$'n'$$ is an integer between $$0$$ and $$21$$, then the minimum value of $$n!\left( 21-n \right) !$$ is

A
$$9!2!$$

B
$$10!11!$$

C
$$20!$$

D
$$21!$$

Solution

To find the minimum value of $$\displaystyle n!\left( 21-n \right) !=21!\frac { n!\left( 21-n \right) ! }{ 21! } =21!\frac { 1 }{ ^{ 21 }{ { C }_{ n } } } =\frac { 21! }{ ^{ 21 }{ { C }_{ n } } } $$
For minimum value, $$\displaystyle ^{ 21 }{ { C }_{ n } }$$ is maximum.
Maximum value of $$\displaystyle ^{ 21 }{ { C }_{ n } }=\frac { ^{ 21 }{ { C }_{ 21-1 } } }{ 2 } =^{ 21 }{ { C }_{ 10 } }$$
$$\therefore$$ Minimum value $$\displaystyle =\frac { 21! }{ ^{ 21 }{ { C }_{ 10 } } } =\frac { 21! }{ 21! } \times 10!11!=10!11!$$
Question 8
1 / -0

If $$^8C_r = ^8C_{r+2}$$, then the value of $$^rC_2$$ is:

A
8

B
3

C
5

D
2

Solution

Since $$ { C }_{ r }^{ n }={ C }_{ n-r }^{ n }$$
$$ \Longrightarrow r=8-(r+2)$$
Hence $$r=3 $$
$$^3C_2=3$$
Question 9
1 / -0

$$^{50}C_{11}+^{50}C_{12}+^{51}C_{13}-^{52}C_{13}=$$

A
$$^{52}C_{14}$$

B
$$0$$

C
$$^{53}C_{13}$$

D
none of these

Solution

We know that
$$\:^{n}C_{r}+\:^{n}C_{r+1}=\:^{n+1}C_{r+1}$$
Application of the above formula gives us
$$\:^{50}C_{11}+\:^{50}C_{12}$$

$$=\:^{51}C_{12}$$

$$\rightarrow \:^{51}C_{12}+\:^{51}C_{13}$$

$$=\:^{52}C_{13}$$

$$\rightarrow \:^{52}C_{13}-\:^{52}C_{13}$$

$$=0$$.
Question 10
1 / -0

The value of $$^{95}C_4+\displaystyle \sum_{j=1}^5 {\;}^{100-j}C_3$$ is

A
$$^{95}C_5$$

B
$$^{100}C_4$$

C
$$^{99}C_4$$

D
$$^{100}C_5$$

Solution

$$^{95}C_4+\displaystyle \sum_{j=1}^5 {\;}^{100-j}C_3$$
$$= ^{95}C_4 + ^{99}C_3 + ^{98}C_3 + ^{97}C_3 + ^{96}C_3 + ^{95}C_3$$
$$= (^{95}C_3 + ^{95}C_4) + ^{96}C_3 + ^{97}C_3 + ^{98}C_3 + ^{99}C_3$$
$$= (^{96}C_4 + ^{96}C_3) + ^{97}C_3 + ^{98}C_3 + ^{99}C_3$$ ($$\because ^{n}C_r+^{n}C_{r-1}=\ ^{n+1}C_r$$)
$$= (^{97}C_4 + ^{97}C_3) + ^{98}C_3 + ^{99}C_3$$ ($$\because ^{n}C_r+^{n}C_{r-1}=\ ^{n+1}C_r$$)
$$= (^{98}C_4 + ^{98}C_3) + ^{99}C_3 $$ ($$\because ^{n}C_r+^{n}C_{r-1}=\ ^{n+1}C_r$$)
$$= ^{99}C_4 + ^{99}C_3 = ^{100}C_4$$ ($$\because ^{n}C_r+^{n}C_{r-1}=\ ^{n+1}C_r$$)

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Re-Attempt Weekly Quiz Competition

Permutations and Combinations Test 12

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Permutations and Combinations Test 12

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SHARING IS CARING

Question 1

11

12

13

14

Solution

Question 2

$$\cfrac{n+1}{2}$$

$$\cfrac{2^{n-1}+1}{n+1}$$

$$\cfrac{2^n}{n}$$

$$\cfrac{n}{2}$$

Solution

Question 3

$$5$$

$$6$$

$$7$$

$$8$$

Solution

Question 4

$$792$$

$$795$$

$$790$$

None of these

Solution

Question 5

125

76

65

100

Solution

Question 6

$$2^{n}$$

$$2^{n\, +\, 1}$$

$$2^{n\, -\, 1}$$

$$2^{2n}$$

Solution

Question 7

$$9!2!$$

$$10!11!$$

$$20!$$

$$21!$$

Solution

Question 8

8

3

5

2

Solution

Question 9

$$^{52}C_{14}$$

$$0$$

$$^{53}C_{13}$$

none of these

Solution

Question 10

$$^{95}C_5$$

$$^{100}C_4$$

$$^{99}C_4$$

$$^{100}C_5$$

Solution

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