Permutations and Combinations Test 31

Result

Permutations and Combinations Test 31

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

If $${ S }_{ n }={ C }_{ 0 }{ C }_{ 1 }+{ C }_{ 1 }{ C }_{ 2 }+.......+{ C }_{ n-1 }{ C }_{ n }$$ and $$\frac { { S }_{ n+1 } }{ { S }_{ n } } =\frac { 15 }{ 4 } $$, then value of n

A
3, 7

B
2, 4

C
1, 3

D
1, 2

Solution
Question 2
1 / -0

If $$x+y=1$$, then $$\displaystyle\sum^n_{r=0}r\cdot {^{n}C_r}x^r\cdot y^{n-r}=?$$

A
$$1$$

B
n

C
nx

D
ny

Solution
Question 3
1 / -0

The number of ways of 3 scholarship of unequal value be awarded to 17 candidates, Such that no candidate gets more than one scholarship is

A
$$^{ 17 }{ C }_{ 3 }$$

B
$${ 17 }^{ 3 }$$

C
$${ 3 }^{ 17 }$$

D
$$^{ 17 }P_{ 3 }$$

Solution

There are $$3$$ scholarships to be awarded between $$17$$ candidates so that no one candidate can get more than one scholarship.

The $$1st$$ scholarship can be given to any of the $$17$$ candidates, So $$17$$ choices.

$$\Rightarrow$$ The $$2nd$$ can be given to the remaining $$16$$ candidates in $${^{16}C_1}=16$$ ways.

Similarly the third in $$15$$ ways.

$$\Rightarrow Total =17\times16\times15={^{17}P_3}$$

Hence, the answer is $${^{17}P_3}$$
Question 4
1 / -0

The number of permutations of letters of the word "PARALLAL" atken four at a time must be,

A
$$216$$

B
$$244$$

C
$$286$$

D
$$1680$$

Solution

Permutation of word PARALLAL taken four at a time
Four letter word might be $$LLL$$_
So total of $$16$$ words [As $$4$$ways, $$4$$spots]
For $$AALL$$, $$6$$ different arrangements
$$AA$$_ _, can be $${ 4 }_{ { C }_{ 2 } }$$ ways $$=6$$ ways
Arranged in $$12$$ ways is total of $$72$$ words
And extra words can be formed in $$120$$ ways
$$16+6+144+120=286$$ ways
Question 5
1 / -0

The value of $$\sum _{ r=0 }^{ s }{ \sum _{ s=1 }^{ n }{ } } ^nC_s.$$ $$^sC_r$$ (where $$r\le s$$ is )

A
$$3^n$$

B
$$n.3^{n-1}$$

C
$$n.3^{n-1}-1$$

D
$$3^n-1$$

Solution
Question 6
1 / -0

There are n identical red balls & m identical green balls. The number of different linear arrangements consisting of "n red balls but not necessarily all the green balls" is $${^{x}C_y}$$ then?

A
$$x=m+n, y=m$$

B
$$x=m+n+1, y=m$$

C
$$x=m+n+1, y=m+1$$

D
$$x=m+n. y=n$$

Solution
Question 7
1 / -0

The exponent of 11 in $$^{200}C_{125}$$ is

A
2

B
1

C
4

D
5

Solution
Question 8
1 / -0

$$1+1.1!+2.2!+3.3!+...+n.n!$$ is equal to

A
$$n!$$

B
$$(n-1)!$$

C
$$(n+1)!$$

D
$$n$$

Solution

given,

$$1+1.1!+2.2!+3.3!+......+n.n!$$

$$=1+(2-1).1!+(3-1)!.2!+.......+(n+1-1).n!$$

$$=1+2.1!-1.1!+3.2!-1.2!+...+(n+1)n!-1n!$$

$$=1+2!-1!+3!-2!+4!-3!+......(n+1)!-n!$$

$$=1+(n+1)!-1$$

$$=(n+1)!$$
Question 9
1 / -0

If $$\displaystyle \frac{1}{{^4{C_n}}} = \frac{1}{{^5{C_n}}} + \frac{1}{{^6{C_n}}}$$, then $$n=$$

A
$$3$$

B
$$2$$

C
$$1$$

D
$$0$$

Solution

$$\dfrac { 1 }{ { 4 }_{ { C }_{ n } } } =\dfrac { 1 }{ { 5 }_{ { C }_{ n } } } +\dfrac { 1 }{ { 6 }_{ { C }_{ n } } } $$

$$\Rightarrow \dfrac { 1 }{ \dfrac { 4! }{ \left( 4-n \right) !n! } } =\dfrac { 1 }{ \dfrac { 5! }{ \left( 5-n \right) !n! } } +\dfrac { 1 }{ \dfrac { 6! }{ \left( 6-n \right) !n! } } $$

$$\Rightarrow \dfrac { \left( 4-n \right) ! }{ 4! } =\dfrac { \left( 5-n \right) ! }{ 5! } +\dfrac { \left( 6-n \right) ! }{ 6! } $$

$$\Rightarrow \dfrac { 1 }{ 4! } =\dfrac { 5-n }{ 5! } +\dfrac { \left( 6-n \right) \left( 5-n \right) }{ 6! } $$

$$\Rightarrow \dfrac { 5-n }{ 5 } +\dfrac { \left( 6-n \right) \left( 5-n \right) }{ 6\times 5 } =1$$

$$\Rightarrow \left( 5-n \right) \left( \dfrac { 1 }{ 5 } +\dfrac { 6-n }{ 30 } \right) =1$$

$$\Rightarrow \left( 5-n \right) \left[ \dfrac { 12-n }{ 30 } \right] =1$$

$$\Rightarrow \left( 5-n \right) \left( 12-n \right) =30$$

$$\Rightarrow { n }^{ 2 }-17n+30=0$$

$$\Rightarrow n=15$$ or $$n=2$$ $$\because$$ $$n$$ cannot be greater than $$4$$

$$\therefore$$ $$n=2$$ [B]
Question 10
1 / -0

The number of such numbers which are even (all digits are different) is

A
$$60$$

B
$$96$$

C
$$120$$

D
$$204$$