Permutations and Combinations Test 47

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Permutations and Combinations Test 47

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Question 1
1 / -0

The value of $$\sum^{10}_{r=0}\begin{pmatrix}10\\r\end{pmatrix}\begin{pmatrix}15\\14-r\end{pmatrix}$$ is equal to

A
$$^{25}C_{12}$$

B
$$^{25}C_{15}$$

C
$$^{25}C_{10}$$

D
$$^{25}C_{14}$$

Solution
Question 2
1 / -0

Two lines intersect at $$O$$. Points $$A_{i}$$ and $$B_{i} (i = 1, 2, ...., n)$$ are taken on these two lines respectively, the number of triangles that can be drawn with the help of these $$2n + 1$$ points is

A
$$n$$

B
$$n^{2}$$

C
$$n^{3}$$

D
None of these

Solution

Number of triangles $$=$$ Taking any two points in $$B_{1}, B_{2}, ..., B_{n}$$ and one point from $$A_{1}, A_{2}, ...., A_{n}$$ or taking any two points from $$A_{1}, A_{2}, A_{3}, ....A_{n}$$ and one point from $$B_{1}, B_{2}, ..., B_{n}$$ or one point $$O$$ and one point from $$A_{1}, A_{2} ....A_{n}$$ and one point from $$B_{1}, B_{2}, ..., B_{n}$$
$$\therefore$$ Required ways
$$= ^{n}C_{2}\times ^{n}C_{1} + ^{n}C_{1}\cdot ^{n}C_{2} + ^{n}C_{1} \times ^{n}C_{1} \times 1$$
$$= \dfrac {n^{2}(n - 1){2}} + \dfrac {n^{2}(n - 1)}{2} + n^{2}$$
$$= (n^{2}(n - 1) + n^{2}(n - 1) + 2n^{2})/2$$
$$= n^{2} [2n - 2 + 2]/2 = n^{3}$$
Question 3
1 / -0

Value of $$\sum _{ k=0 }^{ n }{ { _{ }^{ k }{ C } }_{ n }\sin { \left( kx \right) } \cos { \left( n-k \right) } } $$ is

A
$${ 2 }^{ n-1 }\left( \sin { nx } \right) $$

B
$${ 2 }^{ n }\sin { \left( nx \right) } $$

C
$${ 2 }^{ n }\cos { \left( nx \right) } $$

D
none of these

Solution
Question 4
1 / -0

In a triangle $$ABC$$, the value of the expression $$\displaystyle \sum_{r = 0}^{n}\ ^{n}C_{r}a^{r}.b^{n - r}.\cos (rB - (n - r)A)$$ is equal to

A
$$c^{n}$$

B
zero

C
$$a^{n}$$

D
$$b^{n}$$

Solution
Question 5
1 / -0

Seven person $$P_1,P_2......, P_7$$ initially seated at chairs $$C_1,C_2,.....C_7$$ respectively.They all left there chairs simultaneously for hand wash. Now in how many ways they can again take seats such that no one sits on his own seat and $$P_1$$, sits on $$C_2$$ and $$P_2$$ sits on $$C_3$$ ?

A
$$52$$

B
$$53$$

C
$$54$$

D
$$55$$

Solution
Question 6
1 / -0

If $$m$$ denotes the number of $$5$$ digit numbers if each successive digits are in their descending order of magnitude and $$n$$ is the corresponding figure. When the digits and in their ascending order of magnitude then $$(m-n)$$ has the value

A
$$ { _{ }^{ 9 }{ C } }_{ 4 }$$

B
$${ _{ }^{ 9 }{ C } }_{ 5 }$$

C
$${ _{ }^{ 10 }{ C } }_{ 3 }$$

D
$$ { _{ }^{ 9 }{ C } }_{ 3 }$$

Solution

$$ {\textbf{Step 1: Find m}} $$

$$ {\text{For m,}} $$

$$ {\text{First we select any 5 digits from 0,1,2,}}...{\text{,9}} $$

$$ {\text{Number of ways = }}{}^{10}{{\text{C}}_5} $$

$$ {\text{Now after selection there is only 1 way to arrange these selected digits, i}}{\text{.e}}{\text{., in descending order}}{\text{.}} $$

$$ {\text{Therefore m = }}{}^{10}{{\text{C}}_5}\times{\text{ 1 = }}{}^{10}{{\text{C}}_5} $$

$$ {\textbf{Step 2: Find n}} $$

$$ {\text{For n,First we select any 5 digits from 1,2,}}...{\text{,9}} $$

$$ {\text{We can't select zero as first digit because then the number won't be a 5 - digit number}}{\text{.}} $$

$$ {\text{Therefore number of ways = }}{}^9{{\text{C}}_5} $$

$$   \Rightarrow {\text{n = }}{}^9{{\text{C}}_5}\times{\text{ 1 = }}{}^9{{\text{C}}_5} $$

$$   \Rightarrow {\text{m - n = }}{}^{10}{{\text{C}}_5}{\text{ - }}{}^9{{\text{C}}_5} $$

$$ {\text{We know that,}}{}^n{{\text{C}}_r}{\text{ + }}{}^n{{\text{C}}_{r - 1}}{\text{ = }}{}^{n + 1}{{\text{C}}_r} $$

$$   \Rightarrow {}^{n + 1}{{\text{C}}_r}{\text{ - }}{}^n{{\text{C}}_r}{\text{ = }}{}^n{{\text{C}}_{r - 1}} $$

$$   \Rightarrow {}^{10}{{\text{C}}_5}{\text{ - }}{}^9{{\text{C}}_5}{\text{ = }}{}^9{{\text{C}}_4} $$

$$ {\text{Hence, m - n = }}{}^9{{\text{C}}_4} $$

$$ {\textbf{Hence, the correct answer is option A}} $$
Question 7
1 / -0

There are $$2$$ identical white balls, $$3$$ identical red balls and $$4$$ green balls of different shades. The number of ways in which they can be arranged in a row so that atleast one ball is separated from the balls of the same colour, is

A
$$6(7! - 4!)$$

B
$$7(6! - 4!)$$

C
$$8! - 5!$$

D
None

Solution

These are totally $$9$$ balls of which $$2$$ are identical of one kind, $$3$$ are a like of another kind $$and$$ $$4$$ district ones.

At least one ball of same color separated $$=$$ Total $$-$$ No ball of same color is separated
Total permutation $$=\dfrac{9!}{2!3!}$$

For no ball is separated : we consider all balls of same color as $$1$$ entity, so there are $$3$$ entities which can be placed in $$3!$$ ways.

The white and red balls are identical so they will be placed in $$1$$ way whereas green balls are different so they can be placed in $$4!$$ ways
$$\Rightarrow Req=3!\times 4!$$

At least one ball is separated $$=\dfrac{9!}{2!3!}-3!4!$$

                                                  $$=\dfrac{9\times 8\times 7!}{2\times 6}-6\times 4!$$

$$=6\left(7!\right)-6\left(\times 4!\right) $$

                                                  $$=6\left( 7!-4!\right ).$$

Hence, the answer is $$6\left( 7!-4!\right ).$$
Question 8
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Two classrooms A and B having capacity of $$25$$ and $$(n-25)$$ seats respectively. $$A_n$$ denotes the number of possible seating arrangements of room $$'A'$$, when 'n' students are to be seated in these rooms, starting from room $$'A'$$ which is to be filled up to its capacity. If $$A_n-A_{n-1}=25!(^{49}C_{25})$$ then 'n' equals:

A
$$50$$

B
$$48$$

C
$$49$$

D
$$51$$

Solution

Given $$A_n=nC_{25} \cdot 25!$$

$$A_{n-1}={n-1}C_{25} \cdot 25!$$

Hence $$nC_{25} \cdot 25! - (n-1)C_{25} \cdot 25!=25! 49C_{25}$$

$$\Rightarrow (n-1)C_{25}+(n-1)C_{24}-(n-1)C_{25}=49C_{24}$$

$$\Rightarrow n-1=49$$

$$\Rightarrow n=50$$
Question 9
1 / -0

If $$\displaystyle \sum_{k = 1}^{n = 1} \ ^{n - k}C_{r} = ^{x}C_{y}$$ then

A
$$x = n + 1; y = r$$

B
$$x = n; y = r + 1$$

C
$$x = n; y = r$$

D
$$x = n + 1; y = r + 1$$

Solution
Question 10
1 / -0

If $${ \left( 1+x \right) }^{ n }=\sum _{ r=0 }^{ n }{ { a }_{ r }{ x }^{ r } } $$ and $${ b }_{ r }=1+\cfrac { { a }_{ r } }{ { a }_{ r-1 } } $$ and $$\prod _{ r=1 }^{ n }{ { b }_{ r }=\cfrac { { \left( 101 \right) }^{ 100 } }{ 100! } } $$, then $$n$$ equals to:

A
$$99$$

B
$$100$$

C
$$101$$

D
None of these

Solution