Permutations and Combinations Test 49

Result

Permutations and Combinations Test 49

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Question 1
1 / -0

A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is

A
140

B
196

C
280

D
346

Solution

There would be two cases,

(1) When 4 is selected from first five and rest 6 from remaining 8.
This is done in $$^{5}C_{4}\times ^{8}C_{6}$$ ways.

(2) When all 5 are selected from first five and rest five from remaining 8.
This can be done in $$^{5}C_{5}\times ^{8}C_{5}$$ ways.

$$\therefore $$ A total of $$^{5}C_{4}\times ^{8}C_{6}+^{5}C_{5}\times ^{8}C_{5}$$

$$=5\times \cfrac{8!}{6!2!}+\cfrac{8!}{3!5!}$$

$$=\cfrac{5\times 8\times 7}{2}+\cfrac{8\times 7\times 8}{3\times 2}$$

$$=140+56=196\,ways$$
Question 2
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If $$S_n=\displaystyle\sum^{n}_{r=0}\dfrac{1}{^{n}C_r}$$ and $$t_n=\displaystyle\sum^n_{r=0}\dfrac{r}{^{n}C_r}$$, then $$\dfrac{t_n}{S_n}$$ is equal to?

A
$$\dfrac{1}{2}n$$

B
$$\dfrac{1}{2}n-1$$

C
$$n-1$$

D
$$\dfrac{2n-1}{2}$$

Solution
Question 3
1 / -0

$$2C_0 + \dfrac{2^2}{2} C_1 + \dfrac{2^3}{3} C_2 +........+ \dfrac{2^{11}}{11} C_{10}$$ is equal to

A
$$\dfrac{2^{11} -1}{11}$$

B
$$\dfrac{3^{11} -1}{11}$$

C
$$\dfrac{3^{11} -1}{12}$$

D
$$\dfrac{3^{11} +1}{11}$$

Solution
Question 4
1 / -0

The number of 5 letter words formed using letters of word "CALCULUS" is

A
$$280$$

B
$$15$$

C
$$1110$$

D
$$56$$

Solution
Question 5
1 / -0

If $$^{2n}C_{3}:^{n}C_{2}::44:1$$, then the value of $$n$$ is

A
$$17$$

B
$$6$$

C
$$11$$

D
$$none\ of\ these$$

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 are in their ascending order of magnitude then $$(m-n)$$ has the value?

A
$$^{10}C_4$$

B
$$^9C_5$$

C
$$^{10}C_3$$

D
$$^9C_3$$

Solution

$$\textbf{Step-1: For descending order}$$
$$\text{In this case, we can choose 5 numbers from 10 numbers.}$$
$$\text{This can be done in}$$ $$^{10}C_5$$ $$\text{ways.}$$
$$\text{Now, as there is only one way in which five numbers can be arranged,}$$ $$m=^{10}C_5.$$
$$\textbf{Step-2: For ascending order}$$
$$\text{In this case, we have 9 options to choose from, as here, 0 will be in front and the }$$
$$\text{number will be 4-digit.}$$
$$\text{Hence,}$$ $$n=^{9}C_4$$
$$\textbf{Step-3: Finding}$$ $$\mathbf{m-n}$$
$$m-n=^{ 10}C_5-^9C_4$$
$$=^9C_5$$
$$\textbf{Hence, the correct option is (B).}$$
Question 7
1 / -0

$$10$$ IIT and $$2$$ NIT students sit at random in a row, and then number of ways in which exactly $$3$$ IIT students sit between $$2$$ NIT students is

A
$$16\times 10!$$

B
$$15\times 10!$$

C
$$10\times 16!$$

D
$$None\ of\ these$$

Solution
Question 8
1 / -0

$${ if}^{ n }{ c }_{ 10 }{ = }^{ n }{ c }_{ 14 }$$ then the value of n is equal to

A
14

B
24

C
34

D
44

Solution
Question 9
1 / -0

The sum $$^{ n }{ C }_{ 0 }+^{ n }{ C }_{ 1 }+^{ n }{ C }_{ 2 }+.......+^{ n }{ C }_{ n }$$ is equal to

A
$$\frac { 2.4.6.........2n }{ n\quad ! } $$

B
$${ n }^{ n }$$

C
$$n!$$

D
$${ 3 }^{ n }$$

Solution
Question 10
1 / -0

If $$\displaystyle\sum^m_{k=1}(k^2+1)k!=1999(2000!)$$, then m is?

A
$$1999$$

B
$$2000$$

C
$$2001$$

D
$$2002$$

Solution