Permutations and Combinations Test 36

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Permutations and Combinations Test 36

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Weekly Quiz Competition

Question 1
1 / -0

A school committee consists of $$2$$ teachers and $$4$$ students. The number of different committees that can be formed from $$5$$ teachers and $$10$$ students is

A
$$200$$

B
$$2100$$

C
$$2000$$

D
$$3200$$

Solution

Given, a school committee consists of $$2$$ teachers and $$4$$ students.
The number of different committees that can be formed is simply, choosing $$2$$ out of the $$5$$ teachers and $$4$$ students out of the $$10$$ students, we get -

$$\Rightarrow$$ $$n=\,^5C_2\times \,^{10}C_4$$

$$=\dfrac{5\times 4}{2\times 1}\times\dfrac{10\times 9\times 8\times 7}{4\times 3\times 2\times 1}$$

$$=2100$$
Question 2
1 / -0

Number of cyphers at the end of $$^{2002}$$ C$$_{1001}$$ is

A
$$0$$

B
$$1$$

C
$$2$$

D
None of these

Solution
Question 3
1 / -0

If $$\displaystyle \sum _{ r=0 }^{ n }{ \left\{ \dfrac { { n{ C }_{ r-1 } } }{ n{ C }_{ r }+n{ C }_{ r-1 } } \right\} } =2 $$ then n is equal to

A
3

B
4

C
5

D
6

Solution

Given,
$$\displaystyle \sum \limits_{ r=0 }^{ n }{ \left\{ \frac { { ^n{ C }_{ r-1 } } }{ n{ C }_{ r }+n{ C }_{ r-1 } } \right\} } =2 $$

or,  $$\displaystyle \sum\limits _{ r=0 }^{ n }{ \left\{ \frac { { ^n{ C }_{ r-1 } } }{^{( n+1)}{ C }_{ r } } \right\} } =2 $$

or,  $$\sum\limits _{ r=0 }^{ n }\dfrac{\dfrac{n!}{(r-1)!(n-r+1)!}}{\dfrac{(n+1)!}{(n+1-r)!(r)!}} =2$$

or,  $$\sum\limits _{ r=0 }^{ n }{\dfrac{r}{n+1} } =2$$

or, $$\dfrac{1}{n+1}\times \dfrac{n(n+1)}{2}=\dfrac{}{}$$

or, $$\dfrac{n}{2}=2$$

or, $$n=4$$.
Question 4
1 / -0

If $$\frac{3^{3 n} \cdot 2^{n}}{108}+\frac{3^{3 n}}{729}+\frac{3^{3 n} \cdot 2^{2 n}}{48}+\frac{2^{3 n} \cdot 3^{3 n}}{64}=37^{3} \cdot 3^{6}$$
, then find the value of n ?

A
2

B
3

C
4

D
5

E
none of these
Question 5
1 / -0

The number of all the possible selection which a student can make for answering one or more questions out of eight given question in a paper, which each question has an alternative is

A
$$255$$

B
$$6560$$

C
$$6561$$

D
none of these

Solution

No. of ways$$={^{8}C_1}+{^{8}C_2}+{^{8}C_3}+{^{8}C_4}+{^{8}C_5}+{^{8}C_6}+{^{8}C_7}+{^{8}C_8}$$
$$=({^{8}C_0}+{^{8}C_1}+......{^{8}C_8})-{^{8}C_0}$$
$$=2^8-1=256-1$$
$$=255$$.
Question 6
1 / -0

$$^{n }{ C }_{ r }+^{ n }{ C }_{ r+1 }$$ is equal to______________.

A
$$^{ n }{ C }_{ R+1 }$$

B
$$^{ n }{ C }_{ R+1 }$$

C
$$^{ n+1 }{ C }_{ R+1 }$$

D
$$^{ n-1 }{ C }_{ R+1 }$$

Solution

$$^nC_{r} + ^nC_{r+1}$$

$$= \dfrac{n!}{r!(n-r)!} + \dfrac{n!}{(r+1)!(n-r-1)!}$$

$$= \dfrac{(r+1)\times n!}{(r+1)!(n-r)!} + \dfrac{(n-r)n!}{(r+1)!(n-r)!}$$

Taking LCM we get,

$$= \dfrac{(n+1)n!}{(r+1)!(n-r)!}$$

$$= \dfrac{(n+1)!}{(r+1)!((n+1)-(r+1))!}$$

$$=\, ^{n+1}C_{r+1}$$
Question 7
1 / -0

If $$^{n}C_{3} + ^{n}C_{4} > ^{n + 1}C_{3}$$, then

A
$$n + 1$$

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

C
$$n + 2$$

D
None of these

Solution
Question 8
1 / -0

$$\displaystyle\sum^{m}_{r=0}{^{n+r}C_n}$$ is equal to?

A
$$^{n+m+1}C_{n+1}$$

B
$$^{n+m+2}C_n$$

C
$$^{n+m+3}C_{n-1}$$

D
None of these

Solution

since $$^{n}\textrm{C}_r=^{n}\textrm{C}_{n-r}$$
$$^{n}\textrm{C}_{n-r}+^{n}\textrm{C}_{n}=^{n+1}\textrm{C}_{r}$$
$$\sum^{m}_{r=0}{^{n+r}C_n}=\sum^{m}_{r=0}{^{n+r}C_r}=^{n}\textrm{C}_{0}+^{n+1}\textrm{C}_{1}+^{n+2}\textrm{C}_{2} \cdot \cdot \cdot\cdot ^{n+m}\textrm{C}_{m} $$
$$=[1 + (n+1)]+^{n+2}\textrm{C}_{2}+^{n+3}\textrm{C}_{3}\cdot \cdot \cdot \cdot ^{n+m}\textrm{C}_{m}$$
$$=[^{n+2}\textrm{C}_{1}+^{n+2}\textrm{C}_{2}]+^{n+3}\textrm{C}_{3}\cdot \cdot \cdot \cdot ^{n+m}\textrm{C}_{m}$$
$$\because [n+2]=^{n+2}\textrm{C}_{1}$$
$$=[^{n+3}\textrm{C}_{2}+^{n+4}\textrm{C}_{3}]+^{n+4}\textrm{C}_{4}\cdot \cdot \cdot \cdot ^{n+m}\textrm{C}_{m}$$and so on....
$$=^{n+m}\textrm{C}_{m-1}+^{n+m}\textrm{C}_{m-1}$$
$$=^{n+m+1}\textrm{C}_{m}=^{n+m+1}\textrm{C}_{n+1}\Rightarrow [^{n}\textrm{C}_{r}=^{n}\textrm{C}_{n-r}]$$
Required answer
Question 9
1 / -0

The number of permutations which can be formed out of the letters of the word "SERIES" three letters together, is:

A
120

B
60

C
42

D
none

Solution
Question 10
1 / -0

If $$\alpha = ^{m}C_{2}$$, then $$^{\alpha}C_{2}$$ is equal to

A
$$^{m + 1}C_{4}$$

B
$$^{m - 1}C_{4}$$

C
$$3 \ \ ^{m + 2}C_{4}$$

D
$$3 \ \ ^{m + 1}C_{4}$$

Solution

$$\alpha = ^{m}C_{2}\Rightarrow \alpha = \dfrac {m(m -1)}{2}$$

$$\therefore ^{a}C_{2} = \dfrac {\alpha (\alpha - 1)}{2} = \dfrac {1}{2} \dfrac {m(m - 1)}{2} \left \{\dfrac {m(m - 1)}{2} - 1\right \}$$

$$= \dfrac {1}{8} m(m - 1)(m - 2)(m + 1)$$

$$= \dfrac {1}{8} (m + 1)m(m -1)(m - 2) = 3 \ \ ^{m + 1}C_{4}$$.

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Re-Attempt Weekly Quiz Competition

Permutations and Combinations Test 36

Result

Permutations and Combinations Test 36

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Rank

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SHARING IS CARING

Question 1

$$200$$

$$2100$$

$$2000$$

$$3200$$

Solution

Question 2

$$0$$

$$1$$

$$2$$

None of these

Solution

Question 3

3

4

5

6

Solution

Question 4

2

3

4

5

none of these

Question 5

$$255$$

$$6560$$

$$6561$$

none of these

Solution

Question 6

$$^{ n }{ C }_{ R+1 }$$

$$^{ n }{ C }_{ R+1 }$$

$$^{ n+1 }{ C }_{ R+1 }$$

$$^{ n-1 }{ C }_{ R+1 }$$

Solution

Question 7

$$n + 1$$

$$\dfrac {n}{2}$$

$$n + 2$$

None of these

Solution

Question 8

$$^{n+m+1}C_{n+1}$$

$$^{n+m+2}C_n$$

$$^{n+m+3}C_{n-1}$$

None of these

Solution

Question 9

120

60

42

none

Solution

Question 10

$$^{m + 1}C_{4}$$

$$^{m - 1}C_{4}$$

$$3 \ \ ^{m + 2}C_{4}$$

$$3 \ \ ^{m + 1}C_{4}$$

Solution

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