Permutations and Combinations Test 33

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Permutations and Combinations Test 33

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Weekly Quiz Competition

Question 1
1 / -0

A shelf contains $$15$$ books, of which $$4$$ are single volume and the others are $$8$$ and $$3$$ volumes respectively. In how many ways can these books be arranged on the shelf so that order of the volumes of same work is maintained $$?$$

A
$$4!$$

B
$$8!$$

C
$$3!$$

D
$$4!8!3!3!$$

Solution

No of books $$15$$
No of ways $$4$$ volumes are arranged are $$4!$$ ways
No of ways $$8$$ volumes are arranged are $$8!$$ ways
No of ways $$3$$ volumes are arranged are $$3!$$ ways
They are arranged in $$3!$$ ways one above the other

So no.of ways is $$4!8!3!3!$$
Question 2
1 / -0

$$\displaystyle \sum^{n-1}_{r=0}\dfrac {^{n}C_{r}}{^{n}C_{r}+^{n}C_{r+1}}=$$

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

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

C
$$(n+1)\dfrac {n}{2}$$

D
$$\dfrac {n\ (n-1)}{2\ (n+1)}$$

Solution

$$\displaystyle \sum^{n-1}_{r=0}\dfrac {^{n}C_{r}}{^{n}C_{r}+^{n}C_{r+1}}$$
we know $$^{n}C_{r}+^{n}C_{r-1}=^{n+1}C_{r}$$
$$\therefore \Rightarrow \sum^{n-1}_{r=0}\dfrac {^{n}C_{r}}{^{n+1}C_{r+1}}$$
Expanding
$$\Rightarrow \sum^{n-1}_{r=0}\frac{n!}{(n-r)!\cdot r!}\cdot \frac{(n-r)!\cdot (r+1)!}{(n+1)!}$$
$$\Rightarrow \sum^{n-1}_{r=0}\frac{r+1}{n+1}$$
$$\Rightarrow \frac{1+2+3\cdot \cdot \cdot n}{n+1}$$
$$\Rightarrow \frac{n}{2}$$
Question 3
1 / -0

If $$a=\,^ { m }C _ { 2 } ,$$ then $$^ { a } C _ { 2 }$$ is equal to

A
$$^{m + 1}C _ { 4 }$$

B
$$ ^{m+2} C _ { 4 }$$

C
$$3.\, ^ { m + 2 } C _ { 4 }$$

D
$$3. \,^ { m + 1 }C_4$$

Solution

$${\textbf{Step-1:Define and give examples of nCr in terms of factorials}}$$
$${\text{Given,}}$$
$$\Rightarrow a=^mC_2$$
$$^aC_2=?$$
$$\Rightarrow a=\dfrac{m!}{(m-2)!2!}$$
$$=\dfrac{m(m-1)}{2!}$$

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

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

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

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

$$=\dfrac{1(m+1)!}{8(m-3)!}$$

$$=\dfrac{3 (m-1)!}{1.2.3.4(m-3)!}$$

$$^aC_2=$$$$3. \,^ { m + 1 }C_4$$
$${\textbf{Hence option D is correct}}$$
Question 4
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$$^{ n }{ C }_{ 1 }.2+^{ n }{ C }_{ 2 }.\frac { { 2 }^{ 2 } }{ 3 } +^{ n }{ C }_{ 3 }.\frac { { 2 }^{ 3 } }{ { 3 }^{ 2 } } +......^{ n }{ C }_{ n }.\frac { { 2 }^{ n } }{ { 3 }^{ n-1 } } =$$

A
$$\frac { { 3 }^{ n }-{ 2 }^{ n } }{ { 3 }^{ n-1 } } $$

B
$$\frac { { 3 }^{ n }+{ 2 }^{ n } }{ { 3 }^{ n-1 } } $$

C
$$\frac { { 5 }^{ n }-{ 3 }^{ n } }{ { 3 }^{ n-1 } } $$

D
$$\frac { { 3 }^{ n }+{ 5 }^{ n } }{ { 3 }^{ n-1 } } $$
Question 5
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The number of seven letter words that can be formed by using the letters of the word $$SUCCESS$$ that the two $$C$$ are together but no two $$S$$ are together is

A
$$24$$

B
$$18$$

C
$$54$$

D
none of these

Solution

using the letters of the word $$SUCCESS$$ that the two $$C$$ are together but no two $$S$$ are together
let two C's be 1 unit
$$\therefore $$ no. of ways $$=\frac{6!}{4!}=30$$
We have to put 1 letter between one S
So, No of ways$$=2\times 3!=12$$
No. of ways=30-12=18
Question 6
1 / -0

Nine boys and 3 girls are to be seated in 2 vans, each having numbered seats, 3 in front and 4 at back. The number of ways of seating arrangements, if the girls should sit together in a back row on adjacent seats, is

A
$$12!$$

B
$$3\times 11!$$

C
$$4\times 11!$$

D
$$3\times 9!$$

Solution
Question 7
1 / -0

Six people are going to sit in a row on a bench. $$A$$ and $$B$$ are adjacent, $$C$$ does not want to sit adjacent to $$D.E$$ and $$F$$ can sit anywhere. Number of ways in which these six people can be seated is

A
$$200$$

B
$$144$$

C
$$120$$

D
$$56$$

Solution

A, B, C, D, E, F
Consider AB as group so we have AB, C, D, E, F.
We have totally $$5$$
No. of ways$$(w_1)=5!\times 2$$
$$=240$$
Let CD are adjacent now AB, CD, E, F
No. of ways$$(w_2)=4!2!2!$$
$$=96$$
Total no. of ways
$$W=w_1-w_2$$
$$=240-96$$
$$=144$$.
Question 8
1 / -0

How many different words can be formed by jumbling the letters in the word $$MISSISSIPPI$$ in which no two $$S$$ are adjacent ?

A
$$8 \times ^ { 6 } C _ { 4 } \times ^ { 7 } C _ { 4 }$$

B
$$6 \times 7 \times ^ { 8 } C _ { 4 }$$

C
$$6 \times 8 \times ^ { 7 } C _ { 4 }$$

D
$$7\times ^{ { 6 } }{ C }_{ { 4 } }\times ^{ { 8 } }{ C }_{ { 4 } }$$

Solution

We have the word $$MISSISSIPPI$$
In this word there are $$4I,4S,2P\,and\,1M$$
No two $$S$$ should be together,
Hence, we can place $$S$$ at these places
Therefore, the possible number of words is given by
$$\begin{array}{l} \frac { { ^{ 8 }{ C_{ 4 } }.7! } }{ { 4!2! } } \\ =\frac { { ^{ 8 }{ C_{ 4 } }.7.6! } }{ { 4.2! } } \\ ={ 7.^{ 8 } }{ C_{ 4 } }{ .^{ 6 } }{ C_{ 4 } } \\ Hence,\, option\, D\, is\, the\, correct\, answer. \end{array}$$
Question 9
1 / -0

In the expansion of $$\left(x^3 - \dfrac{1}{x^2}\right)^{15}$$, the constant terms is

A
$$^{15}C_6$$

B
$$-{^{15}C_6}$$

C
$$^{15}C_4$$

D
$$-{^{15}C_4}$$

Solution

We have,
$${ \left( { { x^{ 2 } }-\dfrac { 1 }{ { { x^{ 2 } } } } } \right) ^{ 15 } } \\ { T_{ r+1 } }{ =^{ 15 } }{ C_{ r } }{ \left( { -1 } \right) ^{ r } }{ x^{ 15-r } } \\ { =^{ 15 } }{ C_{ r } }{ \left( { -1 } \right) ^{ r } }{ x^{ 45-5r } } $$

$$\\ For\, constant\ term\, \\ 45-5t=0 \\ r=9 $$

Therefore,
$$=- ^{ 15 }{ C_{ 9 } } = -^{ 15 }{ C_{ 6 } }$$

Hence, this is the answer.
Question 10
1 / -0

The value of $$^{47}C_{4}+\displaystyle \sum _{ j=1 }^{ 5 }\ ^{ \left( 52-j \right) } { C }_{ 3 }$$ is

A
$$^{47}C_{5}$$

B
$$^{52}C_{5}$$

C
$$^{52}C_{4}$$

D
$$^{52}C_{3}$$

Solution

We have,
$$\begin{array}{l} ^{ 47 }{ C_{ 4 } }{ +^{ 51 } }{ C_{ 3 } }{ +^{ 50 } }{ C_{ 3 } }{ +^{ 49 } }{ C_{ 3 } }{ +^{ 48 } }{ C_{ 3 } }{ +^{ 47 } }{ C_{ 3 } } \\ { =^{ n } }{ C_{ r } }{ +^{ n } }{ C_{ r-1 } }{ =^{ n+1 } }{ C_{ r } } \\ { =^{ 48 } }{ C_{ 4 } }{ +^{ 48 } }{ C_{ 3 } }{ +^{ 49 } }{ C_{ 3 } }{ +^{ 50 } }{ C_{ 3 } }{ +^{ 51 } }{ C_{ 3 } } \\ { =^{ 49 } }{ C_{ 4 } }{ +^{ 49 } }{ C_{ 3 } }{ +^{ 50 } }{ C_{ 3 } }{ +^{ 51 } }{ C_{ 3 } } \end{array}$$
Similarly,
$$=^{52}{C_4}$$.

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Re-Attempt Weekly Quiz Competition

Permutations and Combinations Test 33

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Permutations and Combinations Test 33

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SHARING IS CARING

Question 1

$$4!$$

$$8!$$

$$3!$$

$$4!8!3!3!$$

Solution

Question 2

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

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

$$(n+1)\dfrac {n}{2}$$

$$\dfrac {n\ (n-1)}{2\ (n+1)}$$

Solution

Question 3

$$^{m + 1}C _ { 4 }$$

$$ ^{m+2} C _ { 4 }$$

$$3.\, ^ { m + 2 } C _ { 4 }$$

$$3. \,^ { m + 1 }C_4$$

Solution

Question 4

$$\frac { { 3 }^{ n }-{ 2 }^{ n } }{ { 3 }^{ n-1 } } $$

$$\frac { { 3 }^{ n }+{ 2 }^{ n } }{ { 3 }^{ n-1 } } $$

$$\frac { { 5 }^{ n }-{ 3 }^{ n } }{ { 3 }^{ n-1 } } $$

$$\frac { { 3 }^{ n }+{ 5 }^{ n } }{ { 3 }^{ n-1 } } $$

Question 5

$$24$$

$$18$$

$$54$$

none of these

Solution

Question 6

$$12!$$

$$3\times 11!$$

$$4\times 11!$$

$$3\times 9!$$

Solution

Question 7

$$200$$

$$144$$

$$120$$

$$56$$

Solution

Question 8

$$8 \times ^ { 6 } C _ { 4 } \times ^ { 7 } C _ { 4 }$$

$$6 \times 7 \times ^ { 8 } C _ { 4 }$$

$$6 \times 8 \times ^ { 7 } C _ { 4 }$$

$$7\times ^{ { 6 } }{ C }_{ { 4 } }\times ^{ { 8 } }{ C }_{ { 4 } }$$

Solution

Question 9

$$^{15}C_6$$

$$-{^{15}C_6}$$

$$^{15}C_4$$

$$-{^{15}C_4}$$

Solution

Question 10

$$^{47}C_{5}$$

$$^{52}C_{5}$$

$$^{52}C_{4}$$

$$^{52}C_{3}$$

Solution

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