Permutations and Combinations Test 13

Result

Permutations and Combinations Test 13

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Weekly Quiz Competition

Question 1
1 / -0

The number of ways in which ten candidates $$A_1, A_2,......A_{10}$$ can be ranked such that $$A_1$$ is always above $$A_{10}$$ is

A
$$5!$$

B
$$2(5!)$$

C
$$10!$$

D
$$\dfrac{1}{2}(10!)$$

Solution

Total number of ways to arrange $$A_1,A_2 .. A_{10} = 10!$$
In $$\dfrac{10! }{ 2}$$ combinations $$A_1$$ will be above $$A_2$$ and in the other half $$A_2$$ will be above $$A_1$$.

Ans : $$\dfrac{10! }{2}$$
Question 2
1 / -0

Select a figure from the options which will replace the question mark to complete the given series.

A

B

C

D

Solution

Each element moves on step forward anticlockwise direction.
Option B.
Question 3
1 / -0

In how many ways can $$6$$ girls be seated in a circle?

A
$$6$$

B
$$6!$$

C
$$5!$$

D
$$2\times 5!$$

Solution

$$6$$ girls can be seated in a circle in $$(6-1)!$$ ways
$$=5!=120$$ ways
Question 4
1 / -0

Select the missing number from the given matrix:
5 2 4
4 4 7
2 5 3
18 30 ?

A
$$43$$

B
$$42$$

C
$$33$$

D
$$32$$

Solution

In first column $$(5+4)\times 2=18$$ In second column $$(2+4)\times 5=30$$

so in third column ans

$$=(4+7)\times3$$

$$=11\times 3=33.$$
Question 5
1 / -0

From a well shuffled pack of $$52$$ playing cards two cards drawn at random. The probability that either both are red or both are kings is:

A
$$\dfrac{{\left( {^{26}{C_2} + \,{\,^4}{C_2}} \right)}}{{^{52}{C_2}}}$$

B
$$\dfrac{{\left( {^{26}{C_2} + {\,^4}{C_2} - {\,^2}{C_2}} \right)}}{{^{52}{C_2}}}$$

C
$$\dfrac{{^{30}{C_2}}}{{^{52}{C_2}}}$$

D
$$\dfrac{{^{39}{C_2}}}{{^{52}{C_2}}}$$

Solution

Assuming, cards are drawn without replacement:

Total possible events $$=^{52}C_2$$

P(both red) $$=\dfrac{ ^{26}C_2}{^{52}C_2}$$

P(both king) $$=\dfrac{^{4}C_2}{^{52}C_2}$$

P(both red & king) $$=\dfrac {^2C_2}{{^{52}C_2}}$$

We use the basic addition rule,

P(both black or both queens) = P(both red) + P(both queens) - P(both black as well as queens)

Required probability $$=\dfrac{^{26}C_2+^{4}C_2-^2C_2}{^{52}C_2}$$
Question 6
1 / -0

$$A$$ is twice as fast as $$B$$ is thrice as fast as $$C$$. The journey covered by $$C$$ in $$42$$ minutes, what will be covered by $$A$$ is

A
$$21$$ Min

B
$$64$$ Min

C
$$17$$ Min

D
$$40$$ Min

Solution

time taken ny C to complete the journey $$=$$ 42 minutes
time taken by A to complete the journey $$=\dfrac{42}{2}$$ minutes ($$\because $$ A is twice as fast as C)
$$=21 $$ minutes.
Question 7
1 / -0

If $$^{n + 1}{C_3} = 4\,{\,^n}{C_2}$$ then $$n=$$

A
$$12$$

B
$$10$$

C
$$16$$

D
$$11$$

Solution

Given that,$$^{n+1}{{C}_{3}}={{4.}^{n}}{{C}_{2}}$$

Then $$n=?$$

$$ ^{n+1}{{C}_{3}}={{4.}^{n}}{{C}_{2}} $$

$$ \Rightarrow \dfrac{\left( n+1 \right)!}{3!\left( n+1-3 \right)!}=4.\dfrac{n!}{2!\left( n-2 \right)!} $$

$$ \Rightarrow \dfrac{\left( n+1 \right)n!}{3\times 2!\left( n-2 \right)!}=4.\dfrac{n!}{2!\left( n-2 \right)!} $$

$$ \Rightarrow \dfrac{\left( n+1 \right)}{3}=4 $$

$$ \Rightarrow n+1=12 $$

$$ \Rightarrow n=12-1=11\, $$

Hence, this is the answer.
Option (D) is correct.
Question 8
1 / -0

If the letter of the word $$LATE$$ be permuted and the words so formed be arranged as in a dictionary . Then the rank of $$LATE$$ is :

A
$$12$$

B
$$13$$

C
$$14$$

D
$$15$$

Solution

LATE
No. of words that start with 'A'
$$\Rightarrow 3!=6$$
No. of words that start with E
$$\Rightarrow 3!=6$$
$$6+6=12$$
$$13^{th}$$ word starts with L
$$13^{th}$$ word = LAET
$$14^{th}$$ word = LATE
$$\therefore $$ Rank of LATE is $$14$$
Question 9
1 / -0

Choose the most appropriate option which fits this pattern

A

B

C

D
Question 10
1 / -0

Solve:$$\dfrac{2}{2}+\dfrac{3}{3}+\dfrac{4}{4}+$$...... + upto $$1000$$ terms= ?

A
$$1000$$

B
twice of $$500$$

C
four times of $$250$$

D
All of these

Solution

All the terms simplify to $$1$$ which is being added $$1000$$ times.
Therefore, the final answer is $$1000$$ and it can be seen that all the options are correct.