Permutations and Combinations Test 10

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Permutations and Combinations Test 10

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Question 1

1 / -0

The maximum number of points of intersection of five lines and four circles is

$$60$$

$$72$$

$$62$$

None of these

Solution

Two circles intersect at two distinct points. Two straight lines intersect at one point.
One circle and one straight line intersect at two distinct points.
Then the total numbers of points of intersections are as follows:

Number of ways of selection	Points of intersection
Two straight lines: $$^5C_2$$	$$^5C_2\times 1=10$$
Two circle: $$^4C_2$$	$$^4C_2\times 2=12$$
One line and one circle: $$^5C_1\times ^4C_1$$	$$^5C_1\times ^4C_1\times 2=40$$
Total	$$= 62$$

Question 2
1 / -0

From a group of persons the number of ways of selecting 5 persons is equal to that of 8 persons. The number of persons in the group is

A
13

B
40

C
18

D
21

Solution

we know, number of ways of selecting r persons from n persons =$$^{n}\textrm{C}_{r}$$
i.e. $$^{n}\textrm{C}_{5}$$= $$^{n}\textrm{C}_{8}$$
but, $$^{n}\textrm{C}_{r}$$=$$^{n}\textrm{C}_{n-r}$$
=> $$n=r+(n-r)$$
=> $$n=5+8$$
=> $$n=13$$
Question 3
1 / -0

If $$^nC_{r-1} = 36, \space ^nC_r = 84$$ and $$^nC_{r+1} = 126,$$ then find $$r$$.

A
2

B
3

C
4

D
5

Solution

Here
$$\displaystyle\frac{^nC_r}{^nC_{r-1}} = \displaystyle\frac{84}{36}$$
$$\Rightarrow

\displaystyle\frac{n-r+1}{r} = \displaystyle\frac{7}{3} \quad\quad

\left(\because \quad \displaystyle\frac{^nC_r}{^nC_{r-1}}

= \displaystyle\frac{n-r+1}{r} \right)$$
$$\Rightarrow 3n-3r+3 = 7r$$
$$\Rightarrow 10r - 3n = 3 \quad\quad (i)$$
And
$$\displaystyle\frac{^nC_{r+1}}{^nC_r} =

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

= \displaystyle\frac{126}{84} \quad\quad \left(\because

\quad \displaystyle\frac{^nC_r}{^nC_{r-1}}

= \displaystyle\frac{n-r+1}{r} \right)$$
$$\Rightarrow \displaystyle\frac{n-r}{r+1} = \displaystyle\frac{3}{2}$$
$$\Rightarrow 2n - 2r = 3r + 3$$
$$\Rightarrow 5r - 2n = -3$$
$$\Rightarrow 10r-4n = -6 \quad\quad ...(ii)$$
Subtracting $$(ii)$$ from $$(i)$$, we get $$n = 9$$
from $$(i),\quad 10r - 27 = 3$$
$$\Rightarrow 10r = 30$$
$$\therefore \quad\quad r = 3$$.
Question 4
1 / -0

If $$\displaystyle ^nC_{r-1}=56$$, $$\displaystyle ^nC_r=28$$ and $$\displaystyle ^nC_{r+1}=8$$ then r is equal to

A
8

B
6

C
5

D
None of these

Solution

From the given information, $$\dfrac{\:^{n}C_{r-1}}{\:^{n}C_{r}}=2$$
$$\Rightarrow \dfrac{n!r!(n-r)!}{(r-1)!(n-(r-1))!.n!}=2$$

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

$$\Rightarrow r=2n+2-2r$$
$$\Rightarrow 3r=2n+2$$...(i)

Also, $$\dfrac{\:^{n}C_{r}}{\:^{n}C_{r+1}}=\dfrac{7}{2}$$
$$\Rightarrow \dfrac{n!(n-(r+1))!.(r+1)!}{n!.r!.(n-r)!}=\dfrac{7}{2}$$

$$\Rightarrow \dfrac{r+1}{n-r}=\dfrac{7}{2}$$

$$\Rightarrow 2r+2=7n-7r$$
$$\Rightarrow 9r=7n-2$$ ...(ii)
$$\Rightarrow n=8$$
Hence,
$$\Rightarrow 9r=56-2$$
$$\Rightarrow 9r=54$$
$$\Rightarrow r=6$$
Question 5
1 / -0

The number of numbers of 9 different nonzero digits such that all the digits in the first four places are less than the digit in the middle and all the digits in the last four places are greater than that in the middle is

A
$$\displaystyle 2(4!)$$

B
$$\displaystyle (4!)^2$$

C
$$8!$$

D
none of these

Solution

$$\textbf{Step 1: Given conditions and finding the middle digit.}$$
$$\text{Given:}$$
$$\text{Numbers of 9 different nonzero digits with following conditions}$$
$$\text{Condition 1:Digits in first four places should be less than the middle digit.}$$
$$\text{Condition 2:Digits in last four places should be greater than the middle digit.}$$
$$\text{According to the conditions the middle digit will be 5; first four digits will be 1,2,3,4 and the last four}$$
$$\text{digits will be 6,7,8,9.}$$

$$\textbf{Step 2: Permutations of first 4 digits and last 4 digits.}$$
$$\text{The first four digits can be permutated with 1,2,3,4}$$
$$\text{Hence the number of Permutations will be}$$ $${4!}$$
$$\text{The last four digits can be permutated with 6,7,8,9}$$
$$\text{Hence the number of Permutations will be}$$ $${4!}$$

$$\textbf{Step 3: Final calculation.}$$
   $$\text{Therefore the number of numbers of 9 different nonzero digits is}$$
$$\Rightarrow \text{4!.4!}$$
$$\Rightarrow {(4!)^2}$$

$$\textbf{Hence option B is correct.}$$
Question 6
1 / -0

In a packet there are m different books, n different pens and p different pencils. The number of selections of at least one article of each type from the packet is

A
$$\displaystyle 2^{m+n+p}-1$$

B
$$\displaystyle \left ( m+1 \right )\left ( n+1 \right )\left ( p+1 \right )-1$$

C
$$\displaystyle 2^{m+n+p}$$

D
$$(2^{m}-1)(2^{n}-1)(2^{p}-1)$$

Solution

Since each of $$(m+n+p)$$ items has 2 choices of being selected or not selected; hence total no. of ways of selecting items $$=2\times 2\times 2\times 2\times .................(m+n+p)times={ 2 }^{ m+n+p }$$
but one case in this will be when none of the items is selected.
Therefore, no. of ways of selecting at least one item of each type
$$(2^{m}-1)(2^{n}-1)(2^{p}-1)$$
Hence, option 'D' is correct.
Question 7
1 / -0

The value of $$\displaystyle \sum_{r= 0}^{n}$$ $$\displaystyle ^{n+r}C_{r}$$ is equal to

A
$$\displaystyle ^{2n+1}C_{n}$$

B
$$\displaystyle ^{2n}C_{n-1}$$

C
$$\displaystyle ^{2n}C_{n+1}$$

D
$$\displaystyle ^{2n+1}C_{n-1}$$

Solution

The given series can be written as
$$^{ n }{ C }_{ 0 }+^{ n+1 }{ C }_{ 1 }+^{ n+2 }{ C }_{ 2 }+.....+^{ 2n }{ C }_{ n }\\ =\left\{ ^{ n+1 }{ C }_{ 0 }+^{ n+1 }{ C }_{ 1 } \right\} +^{ n+2 }{ C }_{ 2 }+.....+^{ 2n }{ C }_{ n }\\ =\left\{ ^{ n+2 }{ C }_{ 1 }+^{ n+2 }{ C }_{ 2 } \right\} +.....+^{ 2n }{ C }_{ n }\\ =\left\{ ^{ n+3 }{ C }_{ 2 }+^{ n+3 }{ C }_{ 3 } \right\} +^{ n+4 }{ C }_{ 4 }+....+^{ 2n }{ C }_{ n }\\ =^{ n+4 }{ C }_{ 3 }+^{ n+4 }{ C }_{ 4 }+....+^{ 2n }{ C }_{ n }$$
Proceeding the same way finally we get the sum as
$$^{ 2n+1 }{ C }_{ n }$$
Question 8
1 / -0

For $$\displaystyle 2\leq r\leq n,\binom{n}{r}+2\binom{n}{r-1}+\binom{n}{r-2}$$ is equal to

A
$$\displaystyle \binom{n+1}{r-1}$$

B
$$\displaystyle 2\binom{n+1}{r+1}$$

C
$$\displaystyle 2\binom{n+2}{r}$$

D
$$\displaystyle \binom{n+2}{r}$$

Solution

$$\:^{n}C_{r}+\:^{n}C_{r-1}+\:^{n}C_{r-1}+\:^{n}C_{r-2}$$
$$=(\:^{n}C_{r}+\:^{n}C_{r-1})+(\:^{n}C_{r-1}+\:^{n}C_{r-2})$$
$$=\:^{n+1}C_{r}+\:^{n+1}C_{r-1}$$
$$=\:^{n+2}C_{r}$$
This comes from the property that
$$\:^{n}C_{r-1}+\:^{n}C_{r}=\:^{n+1}C_{r}$$
Question 9
1 / -0

The number of 5-digit even numbers that can be made with the digits 0,1,2 and 3 is

A
384

B
192

C
768

D
none of these

Solution

Here even number must end with $$0$$ or $$2$$

Since $$5$$ digit number is to be formed with $${0,1,2,3}$$ hence it is obvious that repetition is allowed.

$$\left< { a }|{ b }|{ c }|{ d }|{ e } \right> $$
here "$$a$$" has $$3$$ options $${1,2,3}$$ ,$$b,c,d$$ has $$4$$ options

$${0,1,2,3}$$ and "$$e$$" has $$2$$ options $${0} $$ and $${2}$$
Number of $$5$$ digit numbers = $$3\times 4\times 4\times 4\times 2=384$$
Hence, option 'A' is correct.
Question 10
1 / -0

$$\displaystyle ^{n}C_{r+1}+^{n}C_{r-1}+2\times ^{n}C_{r}$$ is equal to

A
$$\displaystyle ^{n+2}C_{r+1}$$

B
$$\displaystyle ^{n+1}C_{r}$$

C
$$\displaystyle ^{n+1}C_{r+1}$$

D
$$\displaystyle ^{n+2}C_{r}$$

Solution

$$\:^{n}C_{r+1}+\:^{n}C_{r-1}+\:^{n}C_{r}+\:^{n}C_{r}$$
$$=(\:^{n}C_{r+1}+\:^{n}C_{r})+(\:^{n}C_{r}+\:^{n}C_{r-1})$$
$$=\:^{n+1}C_{r+1}+\:^{n+1}C_{r}$$
$$=\:^{n+2}C_{r+1}$$
This comes from the property that
$$\:^{n}C_{r-1}+\:^{n}C_{r}=\:^{n+1}C_{r}$$

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Re-Attempt Weekly Quiz Competition

Permutations and Combinations Test 10

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Permutations and Combinations Test 10

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SHARING IS CARING

Question 1

$$60$$

$$72$$

$$62$$

None of these

Solution

Question 2

13

40

18

21

Solution

Question 3

2

3

4

5

Solution

Question 4

8

6

5

None of these

Solution

Question 5

$$\displaystyle 2(4!)$$

$$\displaystyle (4!)^2$$

$$8!$$

none of these

Solution

Question 6

$$\displaystyle 2^{m+n+p}-1$$

$$\displaystyle \left ( m+1 \right )\left ( n+1 \right )\left ( p+1 \right )-1$$

$$\displaystyle 2^{m+n+p}$$

$$(2^{m}-1)(2^{n}-1)(2^{p}-1)$$

Solution

Question 7

$$\displaystyle ^{2n+1}C_{n}$$

$$\displaystyle ^{2n}C_{n-1}$$

$$\displaystyle ^{2n}C_{n+1}$$

$$\displaystyle ^{2n+1}C_{n-1}$$

Solution

Question 8

$$\displaystyle \binom{n+1}{r-1}$$

$$\displaystyle 2\binom{n+1}{r+1}$$

$$\displaystyle 2\binom{n+2}{r}$$

$$\displaystyle \binom{n+2}{r}$$

Solution

Question 9

384

192

768

none of these

Solution

Question 10

$$\displaystyle ^{n+2}C_{r+1}$$

$$\displaystyle ^{n+1}C_{r}$$

$$\displaystyle ^{n+1}C_{r+1}$$

$$\displaystyle ^{n+2}C_{r}$$

Solution

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