Permutations and Combinations Test 23

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Permutations and Combinations Test 23

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Weekly Quiz Competition

Question 1
1 / -0

If $$p> q$$, the number of ways of $$p$$ men and $$q$$ women can be seated in a row so that no two women sit together is

A
$$\cfrac { p!q! }{ \left( p+q \right) ! } $$

B
$$\cfrac { \left( p+q \right) ! }{ p!\left( q+! \right) ! } $$

C
$$\cfrac { p!\left( q+1 \right) ! }{ \left( p-q+1 \right) ! } $$

D
$$\cfrac { p!(p+1)! }{ \left( p-q+1 \right) ! } $$

Solution

Arrangement of $$p$$ men in a row $$=p!$$
$$\therefore$$ Required number of ways $$={ p! }^{ p+1 }{ P }_{ q }=\cfrac { p!(p+1)! }{ \left( p-q+1 \right) ! } $$
Question 2
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Total number of arrangements of the letters of the word SUCCESS such that both $$C's$$ are together and no two $$S's$$ are together is

A
$$12$$

B
$$24$$

C
$$96$$

D
$$120$$

Solution

As $$CC$$ are together and no two $$'S'$$ are together
$$\therefore$$ Consider $$CC$$ as single object. $$U, CC, E$$ can be arranged in $$3!$$ ways
$$\times U\times CC \times E \times$$
Now three $$'S'$$ are to the placed in four available place as $$^{4}c_{3}$$ ways
$$\therefore$$ Required number of ways are $$3! \times ^{4}C_{3} = 24$$.
Hence choice (b) is correct.
Question 3
1 / -0

Out of $$15$$ persons $$10$$ can speak Hindi and $$8$$ can speak English. If two persons are chosen at random, then the probability that one person speaks Hindi only and the other speaks both Hindi and English is

A
$$5/3$$

B
$$7/12$$

C
$$1/5$$

D
$$2/5$$

Solution

$$n(H) = 10$$ {people who speak hindi}
$$n(E) = 8$$ {People who speak English}
$$n(E\cup H) = 15$$ (total)
$$n (E \cap H) = 10 + 8 – 15 = 3$$ (Speaking both)
People who speak hindi only $$= 10 – 3 = 7$$
Total ways of selecting $$2$$ people $$= 15C_{2} = 105$$
Favourable ways $$= 7C_{1} \times 3C_{1} = 21$$
Probability of occurrence $$= \dfrac {21}{105} = \dfrac {1}{5}$$
Question 4
1 / -0

A telephone number $$d_1d_2d_3d_4d_5d_6d_7$$ is called memorable if the prefix sequence $$d_1d_2d_3$$ is exactly the same as either of the sequence $$d_4d_5d_6$$ or $$d_5d_6d_7$$(or possibly both). If each $$d_1\epsilon\{x|0\leq x\leq 9, x\epsilon W\}$$, then number of distinct memorable telephone number is(are).

A
$$19810$$

B
$$19,910$$

C
$$19,990$$

D
$$20,000$$

Solution
Question 5
1 / -0

$$\sum _{ r=0 }^{ n-1 }{ \cfrac { { _{ }^{ n }{ C } }_{ r } }{ { _{ }^{ n }{ C } }_{ r }+{ _{ }^{ n }{ C } }_{ r+1 } } } $$ is equal to

A
$$\cfrac { n }{ 2 } $$

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

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

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

Solution

$$\sum_{r=0}^{n-1}$$$$\dfrac{^nC_r}{^nC_r+^nC_{r+1}}$$

=$$\sum_{r=0}^{n-1}$$$$\dfrac{^nC_r}{^{n+1}C_{r+1}}$$ ----------(because $$^nC_r + ^nC_{r-1}=^{n+1}C_r$$)

=$$\sum_{r=0}^{n-1}\dfrac{^nC_r}{\dfrac{(n+1)}{(r+1)}^nC_r}$$

=$$\sum_{r=0}^{n-1}\dfrac{r+1}{n+1}$$

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

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

hence option $$'A'$$ is correct.
Question 6
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A man $$x$$ has $$7$$ friends, $$4$$ of them are ladies and $$3$$ are men. His wife $$Y$$ also has $$7$$ friends, $$3$$ of them are ladies and $$4$$ are men. Assume $$X$$ and $$Y$$ have no common friends. Then, the total number of ways in which $$X$$ and $$Y$$ together can throw a party inviting $$3$$ ladies and $$3$$ men , so that $$3$$ friends of each of $$X$$ and $$Y$$ are in this party, is

A
$$485$$

B
$$468$$

C
$$469$$

D
$$484$$

Solution

Case 1: One of them invites all three the same gender.
ie, $$N=(^4C_3\ ^4C_3)+(^3C_3\ ^3C_3)=4^2+1^2=17$$

Case 2: One of them invites 1 male and 2 females$$\Rightarrow$$The other invites 2 males and 1 female.
ie, $$N=(^4C_2\ ^3C_1\ ^4C_2\ ^3C_1)+(^4C_1\ ^3C_2\ ^3C_2\ ^4C_1)=18^2+12^2=468 $$

So, Total Possible ways$$=468+17=485$$

ie, Option A is correct answer.
Question 7
1 / -0

The sum $$^{20}C_0+^{20}C_1+^{20}C_2+..... +^{20}C_{10}$$ is equal to

A
$$2^{20}+\dfrac{20 \,!}{(10\,!)^2}$$

B
$$2^{19}+\dfrac{1}{2}.\dfrac{20\,!}{(10\,!)^2}$$

C
$$2^{19}+^{20}C_{10}$$

D
none of these

Solution

Let
$$S={}^{ 20 }{ { C }_{ 0 } }+{}^{ 20 }{ { C }_{ 1 } }+{}^{ 20 }{ { C }_{ 2 } }+.......+{}^{ 20 }{ { C }_{ 10 } }$$
$$\Rightarrow 2S=\left( {}^{ 20 }{ { C }_{ 0 } }+{}^{ 20 }{ { C }_{ 0 } } \right) +\left( {}^{ 20 }{ { C }_{ 1 } }+{}^{ 20 }{ { C }_{ 1 } } \right) +\left( {}^{ 20 }{ { C }_{ 2 }+ }{}^{ 20 }{ { C }_{ 2 } } \right) +.......+\left( {}^{ 20 }{ { C }_{ 9 } }+{}^{ 20 }{ { C }_{ 9 } } \right) +\left( {}^{ 20 }{ { C }_{ 10 } }+{}^{ 20 }{ { C }_{ 10 } } \right) $$
$$\Rightarrow 2S-{}^{ 20 }{ { C }_{ 10 } }=\left( {}^{ 20 }{ { C }_{ 0 } }+{}^{ 20 }{ { C }_{ 20 } } \right) +\left( {}^{ 20 }{ { C }_{ 1 } }+{}^{ 20 }{ { C }_{ 19 } } \right) +\left( {}^{ 20 }{ { C }_{ 2 }+ }{}^{ 20 }{ { C }_{ 18 } } \right) +.......+\left( {}^{ 20 }{ { C }_{ 9 } }+{}^{ 20 }{ { C }_{ 11 } } \right) +{}^{ 20 }{ { C }_{ 10 } }$$
(Using $$ _{ }{}^{ n }{ { C }_{ r } }=_{ }{}^{ n }{ { C }_{ n-r } }$$)
$$\Rightarrow 2S-{}^{ 20 }{ { C }_{ 10 } }={}^{ 20 }{ { C }_{ 0 } }+{}^{ 20 }{ { C }_{ 1 } }+{}^{ 20 }{ { C }_{ 2 }+ }.......+{}^{ 20 }{ { C }_{ 19 } }+{}^{ 20 }{ { C }_{ 20 } }$$
$$\Rightarrow 2S-{}^{ 20 }{ { C }_{ 10 } }={ 2 }{}^{ 20 }$$ (Using $${}^{ n }{ { C }_{ 0 } }+{}^{ n }{ { C }_{ 1 } }+{}^{ n }{ { C }_{ 2 } }+.......+{}^{ n }{ { C }_{ n } }={ 2 }{}^{ n }$$)
$$\Rightarrow S=\dfrac { { 2 }{}^{ 20 }+{}^{ 20 }{ { C }_{ 10 } } }{ 2 } $$
$$\Rightarrow S={ 2 }{}^{ 19 }+\dfrac { 1 }{ 2 } .\dfrac { 20! }{ { (10!) }{}^{ 2 } } $$

Hence, Option B is correct.
Question 8
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The value of $$\begin{pmatrix} 30 \\ 0 \end{pmatrix}\begin{pmatrix} 30 \\ 10 \end{pmatrix}-\begin{pmatrix} 30 \\ 1 \end{pmatrix}\begin{pmatrix} 30 \\ 11 \end{pmatrix}+\begin{pmatrix} 30 \\ 2 \end{pmatrix}\begin{pmatrix} 30 \\ 12 \end{pmatrix}-...+\begin{pmatrix} 30 \\ 20 \end{pmatrix}\begin{pmatrix} 30 \\ 30 \end{pmatrix}$$ is where $$\begin{pmatrix} n \\ r \end{pmatrix}={ _{ }^{ n }{ C } }_{ r }$$

A
$$\begin{pmatrix} 30 \\ 15 \end{pmatrix}$$

B
$$\begin{pmatrix} 60 \\ 15 \end{pmatrix}$$

C
$$\begin{pmatrix} 60 \\ 30 \end{pmatrix}$$

D
$$\begin{pmatrix} 30 \\ 10 \end{pmatrix}$$

Solution

$$\displaystyle^{30}C_{0}\displaystyle^{30}C_{10}-\displaystyle^{30}C_{1}\displaystyle^{30}C_{11}+........+\displaystyle^{30}C_{20}\displaystyle^{30}C_{30}$$
WKT
$$\Rightarrow (x+1)^{30}=\displaystyle^{30}C_{0}+x\left(^{30}C_{1}\right)+x^2\left(^{30}C_{2}\right)+.........+x^{30}\left(^{30}C_{30}\right)$$
$$\Rightarrow (x-1)^{30}=x^{30}\left({^{30}C_{0}}\right)-x^{29}\left({^{30}C_{1}}\right)+......+^{30}C_{30}$$
Multiply the above $$2$$ equations $$\xi$$ comparing $$x^{20}$$ co-efficient.
$$\Rightarrow (x+1)^{30}(x-1)^{30}=\left(x^2-1\right)^{30}$$
$$\Rightarrow T_{10+1}=^{30}C_{10}x^{2(10)}\times 1^{20}={^{30}C_{10}}x^{20}$$
$$\Rightarrow {^{30}C_{10}}={^{30}C_{0}}{^{30}C_{10}}-{^{30}C_{1}}{^{30}C_{11}}+.......+{^{30}C_{20}}{^{30}C_{30}}$$
$$\therefore$$ Required $$={^{30}C_{10}}$$
Hence, the answer is $${^{30}C_{10}}.$$
Question 9
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Let $${ T }_{ n }$$ denotes the number of triangles which can be formed by using the vertices of a regular polygon of n sides. If $${ T }_{ n+1 }\ -\ { T }_{ n }=21$$, then $$n$$ is equal to

A
$$5$$

B
$$7$$

C
$$6$$

D
$$4$$

Solution

Number of triangle that can be formed using polygon of $$n$$ sided $$={}^{ n }{ C }_{3}$$

According to question,

$${}^{ n+1 }{ C }_{3}-{ }^{ n }{ C }_{3}=21$$

$$\dfrac { (n+1)(n)(n-1) }{ 6 } -\dfrac { n(n-1)(n-2) }{ 6 } =21$$

$$\Rightarrow \dfrac { 3(n)(n-1) }{ 6 } =21$$

$$\Rightarrow (n)(n-1)=42$$

Solving this equation we get $$n=-6,7$$

But $$n$$ can't be negative

Hence, $$n=7$$.
Question 10
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If $$15! =2^{\alpha}\cdot 3^{\beta}\cdot 5^{\gamma}\cdot 7^{\delta}\cdot 11^{\theta}\cdot 13^{\Phi}$$, then the value of expression $$\alpha -\beta +\gamma -\delta +\theta -\Phi$$ is

A
$$4$$

B
$$6$$

C
$$8$$

D
$$10$$

Solution

$$15!=1\times 2\times 3\times 4 \times 5\times 6\times 7\times 8\times 9\times 10\times 11\times 12\times 13\times 14\times 15$$

$$15!=2\times 3\times 2^2 \times 5\times 3\times 2 \times 7\times 2^3\times 3^2\times 2\times 5\times 11\times 3\times 2^2 \times 13\times 2\times 7 \times 3\times 5$$

$$15!=2^{11}\times 3^6\times 5^3\times 7^ 2\times 11\times 13 $$

$$\implies \alpha =11$$

$$\implies \beta =6$$

$$\implies \gamma=3$$

$$\implies \delta =2$$

$$\implies \theta =1$$

$$\implies \Phi=1$$

$$\alpha-\beta+\gamma-\delta+\theta-\Phi=11-6+3-2+1-1=6$$

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Permutations and Combinations Test 23

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Permutations and Combinations Test 23

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SHARING IS CARING

Question 1

$$\cfrac { p!q! }{ \left( p+q \right) ! } $$

$$\cfrac { \left( p+q \right) ! }{ p!\left( q+! \right) ! } $$

$$\cfrac { p!\left( q+1 \right) ! }{ \left( p-q+1 \right) ! } $$

$$\cfrac { p!(p+1)! }{ \left( p-q+1 \right) ! } $$

Solution

Question 2

$$12$$

$$24$$

$$96$$

$$120$$

Solution

Question 3

$$5/3$$

$$7/12$$

$$1/5$$

$$2/5$$

Solution

Question 4

$$19810$$

$$19,910$$

$$19,990$$

$$20,000$$

Solution

Question 5

$$\cfrac { n }{ 2 } $$

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

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

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

Solution

Question 6

$$485$$

$$468$$

$$469$$

$$484$$

Solution

Question 7

$$2^{20}+\dfrac{20 \,!}{(10\,!)^2}$$

$$2^{19}+\dfrac{1}{2}.\dfrac{20\,!}{(10\,!)^2}$$

$$2^{19}+^{20}C_{10}$$

none of these

Solution

Question 8

$$\begin{pmatrix} 30 \\ 15 \end{pmatrix}$$

$$\begin{pmatrix} 60 \\ 15 \end{pmatrix}$$

$$\begin{pmatrix} 60 \\ 30 \end{pmatrix}$$

$$\begin{pmatrix} 30 \\ 10 \end{pmatrix}$$

Solution

Question 9

$$5$$

$$7$$

$$6$$

$$4$$

Solution

Question 10

$$4$$

$$6$$

$$8$$

$$10$$

Solution

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