Permutations and Combinations Test 30

Result

Permutations and Combinations Test 30

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

In how many ways atleast one horse and atleast one dog can be selected out of eight horses and seven dogs.

A
$${2}^{15}-2$$

B
$${2}^{15}-1$$

C
$$({2}^{8}-1)({2}^{7}-1)$$

D
$${ _{ }^{ 15 }{ C } }_{ 2 }$$

Solution

As we can either select or not select any horse; total ways for horses$$=2^r$$ ways to be excluded are when no horse is selected, which is only one way, therefore, favourable ways for horses$$=2^8-1$$
similarly, favourable ways for dogs$$=2^7-1$$
$$\Rightarrow$$ Total ways$$=(2^8-1)(2^7-1)$$.
Question 2
1 / -0

Determine $$n$$, if $$^ { 2 n } \mathrm { C } _ { 3 } : ^ { n } \mathrm { C } _ { 3 } = 12 : 1$$

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

Solution

$$^{2n}C_3 : ^nC_3= 12:1$$
first we calculate $$^{2n}C_3$$ and $$^nC_3$$ seperately

$$ ^{2n}C_3 =\dfrac {2n!}{3!(2n-3)!}=\dfrac {2n(2n-1)(2n-2)(2n-3)!}{3\times 2 \times 1 \times (2n-3)!}$$

$$= \dfrac {2n(2n-1)(2n-2)}{6}$$

$$^nC_3 =\dfrac {n!}{3!(n-3)!}= \dfrac {n(n-1)(n-2)(n-3)!}{3\times 2 \times 1 \times (n-3)!}$$

$$ = \dfrac {n(n-1)(n-2)}{6}$$

$$^{2n}C_3 : ^nC_3=12:1$$

$$\therefore \dfrac {2n(2n-1)(2n-2)}{6}:\dfrac {n(n-1)(n-2)}{6}= 12:1$$
$$ \Rightarrow \dfrac {2n(2n-1)(2n-2)}{n(n-1)(n-2)}=12$$

$$\Rightarrow 2\times 2 \dfrac {(2n-1)(n-1)}{(n-1)(n-2)}=12$$
$$\Rightarrow 8n - 4 = 12n - 24$$

$$ \Rightarrow 4n = 20$$
$$\Rightarrow n=\dfrac {20}{4}$$

$$\Rightarrow n = 5$$
Question 3
1 / -0

The total number of ways of arranging the letters $$AAABBBCCDEF$$ in a row such that letters $$C$$ are separated from one another is

A
$$277200$$

B
$$138600$$

C
$$453600$$

D
none of these

Solution
Question 4
1 / -0

$$^{404}C_{4}-^{4}C_{1}^{303}C_{4}+^{4}C_{2}^{202}C_{4}-^{404}C_{4}^{101}C_{4}$$ is equal to

A
$$(401)^{4}$$

B
$$(101)^{4}$$

C
$$0$$

D
$$(201)^{4}$$

Solution
Question 5
1 / -0

The number of positive integral solutions of the equation $$x _ { 1 } x _ { 2 } x _ { 3 } x _ { 4 } x _ { 5 } = 1050$$ is

A
1800

B
1600

C
1400

D
1875

Solution

We have,
$${x_1}{x_2}{x_3}{x_4}{x_5} = 1050 = 2 \times 3 \times {5^2} \times 7$$
Now,
$$2,3$$ and $$7$$ can be put in any boxes $${x_1},{x_2},{x_3},{x_4}$$ and $${x_5}$$.
Also $$5,5$$ can be distributed in $$5$$ boxes in
$$^{2 + 5 - 1}{C_{5 - 1}}{ = ^6}{C_4} = 15$$Ways.
So total no. of positive integral solutions $$=15 \times5 \times5 \times5$$
$$=1875$$
Option $$D$$ is correct answer.
Question 6
1 / -0

Number of different natural numbers which are smaller than two hundred million $$\&$$ using only the digits $$1$$ or $$2$$ is ___________________.

A
$$( 3 ) \cdot 2 ^ { 8 } - 2$$

B
$$( 3 ) \cdot 2 ^ { 8 } - 1$$

C
$$2$$$$\left( 2 ^ { 9 } - 1 \right)$$

D
none

Solution
Question 7
1 / -0

If $$(1+x+x^2)^n=\displaystyle\sum^{2n}_{r=0}a_rx^r$$, then $$a_0a_{2r}-a_1a_{2r+1}+a_2a_{2r+2}-....=?$$

A
$$a_r$$

B
$$a_{n-2r}$$

C
$$a_{n+r}$$

D
$$a_{2r}$$
Question 8
1 / -0

Number of cyphers at the end of 202$$\mathrm { C } _ { 1001 }$$ is ________.

A
0

B
1

C
2

D
None

Solution
Question 9
1 / -0

If $$\displaystyle\sum^{n-r}_{k=1}$$ $$^{n-k}C_r={^{x}C_y}$$ 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

Letters of the word $$MATHEMATICS$$ are arranged in all the possible ways, in how many words letter $$C$$ is between $$S$$ and $$H$$(these three letter are not necessary together)?

A
$$(11!)/(2!)(2!)(2!)$$

B
$$(11!)/(3!)(2!)(2!)$$

C
$$(11!)/(3!)(3!)(2!)$$`

D
$$None\ of\ these$$