Permutations and Combinations Test 51

Result

Permutations and Combinations Test 51

Score
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Rank
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Weekly Quiz Competition

Question 1
1 / -0

There are $$n$$ points on a circle. The number of staight lines formed by joining them is equal to

A
$$^{n}C_{2}$$

B
$$^{n}P_{2}$$

C
$$^{n}C_{2}-1$$

D
$$none\ of\ these$$

Solution
Question 2
1 / -0

In a set of lottery Tickets $$7$$ carry prizes and $$25$$ are blank. If three tickets are drawn then the probability to get a prize is?

A
$$\dfrac{{^{7}C_3}}{{^{32}C_3}}$$

B
$$\dfrac{{^{25}C_3}}{{^{32}C_3}}$$

C
$$1-\dfrac{{^{25}C_3}}{{^{32}C_3}}$$

D
Cannot be decided

Solution
Question 3
1 / -0

The sum of the numbers formed by taking all the digit at a from 0, 2, 3, 4 is

A
57996

B
75669

C
99657

D
57699

Solution
Question 4
1 / -0

Number of six digit numbers whose sum of the digits is $$49$$ are

A
$$^{12}C_{5}$$

B
$$6^{10}$$

C
$$^{10}C_{5}$$

D
$$1200$$
Question 5
1 / -0

The letters of the word 'VICTORY' are arranged in all possible ways, and the words thus obtained are arranged as in a dictionary. Then the rank of given word is

A
3733

B
5309

C
5040

D
3732

Solution
Question 6
1 / -0

$$20$$ soldiers are standing in a row and their captain want to send $$7$$ out of them for a mission. In how many ways can captain select them such that at least one soldier find the soldier next to him is also selected.

A
$$^{20}C_{7}$$

B
$$^{14}C_{7}$$

C
$$^{20}C_{7}-^{13}C_{7}$$

D
$$^{20}C_{7}-^{14}C_{7}$$

Solution

$$\textbf{Step -1: Calculating}$$
$$\text{Required number of ways= (Total number of ways without any restriction )}$$
$$\text{- (No. of ways in which no two consecutive soldiers are selected)}$$
$$\implies \text{Let }x_1\text{ be the number of soldiers selected before the first soldier is selected.}$$
$$\text{Also , let }x_2,x_3,x_4..x_7 \text{ be the number of soldiers between 1st and 2nd, 2nd and 3rd}$$
$$\text{ ...6th and 7th respectively}$$
$$\text{Also let }x_8 \text{ be the number of soldiers after the 7th soldier is selected}$$
$$\textbf{Step -2: Calculating the number of ways .}$$
$$\implies x_1+x_2+...+x_8=13$$
$$x_1,x_8 \geq 0 \text{ and }x_2,x_3..,x_7 \geq 1$$
$$\implies x_1+y_2+...+y_7+x_8=7$$
$$\text{Where }y_2,y_3..,y_7\geq 0$$
$$\text{Number of solution of above equation}$$
$$\implies ^{8+7-1}C_{8-1}= ^ { 14}C_7$$
$$\therefore \text{Required number of ways = } ^{20}C_7-^ {14}C_7$$
$$\textbf{Hence, the required number of ways is }\mathbf{^{20}C_7- ^{14}C_7 .} $$
Question 7
1 / -0

If $$C_{r}$$ stands for $$^{n}C_{r}$$ then the sum of the series $$\frac{2(\frac{n}{2})!(\frac{n}{2})!}{n!}[C_{0}^{2}-2C_{1}^{2}+3C_{2}^{2}-.....+(-1)^{n}(n+1)C_{n}^{2}]$$,
where n is an even positive integer, is

A
$$(-1)^{n/2}(n+2)$$

B
$$(-1)^{n}(n+2)$$

C
$$(-1)^{n/2}(n+1)$$

D
None of these

Solution
Question 8
1 / -0

If $$^{32}C_{2n-1}=^{32}C_{n-3}$$ then $$n=$$

A
$$10$$

B
$$9$$

C
$$12$$

D
$$11$$

Solution
Question 9
1 / -0

The value of $${ C }_{ 0 }^{ 2 }+3{ C }_{ 1 }^{ 2 }+5{ C }_{ 2 }^{ 2 }+.....$$ to $$n+1$$ terms is

A
$$n{ \cdot ^{2n}}{C_n}$$

B
$$(2n+2)\cdot ^{ 2n }{ C }_{ n }$$

C
$$(n + 1){ \cdot ^{2n}}{C_{n - 1}}$$

D
$$(2n + 1){ \cdot ^{2n - 1}}{C_{n - 1}}$$
Question 10
1 / -0

The number of ways is which an examiner can assign $$30$$ marks to $$8$$ questions, giving not less Then $$2$$ marks to any question is-

A
$$^{21}C_{7}$$

B
$$^{21}C_{8}$$

C
$$^{21}C_{9}$$

D
$$^{21}C_{10}$$

Solution