Straight Lines Test 46

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Straight Lines Test 46

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Question 1
1 / -0

There are $$2$$ identical white balls, $$3$$ identical red balls and $$4$$ green balls of different shades. The number of ways in which they can be arranged in a row so that atleast one ball is separated from the balls of the same colour, is

A
$$6(7! - 4!)$$

B
$$7(6! - 4!)$$

C
$$8! - 5!$$

D
None

Solution

These are totally $$9$$ balls of which $$2$$ are identical of one kind, $$3$$ are a like of another kind $$and$$ $$4$$ district ones.

At least one ball of same color separated $$=$$ Total $$-$$ No ball of same color is separated
Total permutation $$=\dfrac{9!}{2!3!}$$

For no ball is separated : we consider all balls of same color as $$1$$ entity, so there are $$3$$ entities which can be placed in $$3!$$ ways.

The white and red balls are identical so they will be placed in $$1$$ way whereas green balls are different so they can be placed in $$4!$$ ways
$$\Rightarrow Req=3!\times 4!$$

At least one ball is separated $$=\dfrac{9!}{2!3!}-3!4!$$

                                                  $$=\dfrac{9\times 8\times 7!}{2\times 6}-6\times 4!$$

$$=6\left(7!\right)-6\left(\times 4!\right) $$

                                                  $$=6\left( 7!-4!\right ).$$

Hence, the answer is $$6\left( 7!-4!\right ).$$
Question 2
1 / -0

If $$^{19}C_r$$ and $$^{19}C_{r-1}$$ are in the ratio 2:3, then find $$^{14}C_r$$

A
91

B
81

C
71

D
61

Solution

We know that, $$ { { n }_{ C } }_{ r }=\dfrac { n! }{ r!(n-r)! } $$

Given, $$ \dfrac { { { 19 }_{ C } }_{ r } }{ { { 19 }_{ C } }_{ r-1 } } =\quad \dfrac { 2 }{ 3 } $$

$$ =>\dfrac { \dfrac { 19! }{ r!(19-r)! } }{ \dfrac { 19! }{ (r-1)!(19-r+1)! } } =\quad \dfrac { 2 }{ 3 } $$

$$ => \dfrac { (r-1)!(20-r)! }{ r!(19-r)! } =\quad \dfrac { 2 }{ 3 } $$

$$ =\dfrac { (r-1)!\times (20-r) \times (19-r)! }{ r\quad \times (r-1)!\times (19-r)! } =\quad \dfrac { 2 }{ 3 } $$
$$ => \dfrac { 20-r }{ r\quad } =\quad \dfrac { 2 }{ 3 } $$
$$ => r = 12 $$

So,
$$ { { 14 }_{ C } }_{ 12 }=\dfrac { 14! }{ 12!(14-12)! } =\dfrac { 14! }{ 12!(2)! } =\dfrac { 14\times 13\times 12! }{ 12!\times 2 } =\quad 7\times 13=91 $$
Question 3
1 / -0

If j, k, and n are consecutive integers such that $$0 < j < k < n$$ and the units (ones) digit of the product jn is 9, what is the units digit of k ?

A
0

B
1

C
2

D
3

E
4

Solution

There are only a few ways you can make a product have a units digit of $$9$$. Either both numbers you’re multiplying have to be $$3$$, or one has to be $$1$$ and one has to be $$9$$. Since we’re dealing with consecutive integers, $$j$$ and $$n$$ can’t both end in $$3$$, so they’re going to have to end in $$1$$ and $$9$$.
It’s important to remember that although we only really care about the units digits in this problem, we’re dealing with numbers that might (or in fact, must) be $$2$$ or more digits. That’s why you can have $$k's$$ units digit be $$0$$.
Say $$j= 19$$, $$k= 20$$, and $$n= 21$$. Then $$jn = (19)(21) = 399$$ (units digit is $$9$$), and the units digit of $$k = 0$$.
Hence option A is correct.
Question 4
1 / -0

Seven person $$P_1,P_2......, P_7$$ initially seated at chairs $$C_1,C_2,.....C_7$$ respectively.They all left there chairs simultaneously for hand wash. Now in how many ways they can again take seats such that no one sits on his own seat and $$P_1$$, sits on $$C_2$$ and $$P_2$$ sits on $$C_3$$ ?

A
$$52$$

B
$$53$$

C
$$54$$

D
$$55$$

Solution
Question 5
1 / -0

How many $$10-digit$$ numbers can be formed by using the digits $$1$$ and $$2$$?

A
$$^{10}{P}_{2}$$

B
$$^{10}{C}_{2}$$

C
$${2}^{10}$$

D
$$10\ !$$

Solution

Each place of a ten digit number can be fixed by any of the two digits. So, the number of ways to form a ten digit number is $${2^{10}}$$.
Question 6
1 / -0

There are infinite, alike, blue, red, green and yellow balls. Find the number of ways to select $$10$$ balls.

A
$$286$$

B
$$84$$

C
$$715$$

D
None of these.

Solution
Question 7
1 / -0

What is the value of $$^nC_n$$?

A
zero

B
$$1$$

C
$$n$$

D
$$n!$$

Solution
Question 8
1 / -0

A
$$6$$

B
$$7$$

C
$$8$$

D
$$9$$

E
$$10$$

Solution
Question 9
1 / -0

In a certain examination paper, there are $$n$$ question. For $$j = 1, 2....n$$, there are $$2^{n-j}$$ students who answered $$j$$ or more questions wrongly. If the total number of wrong answers is $$4095$$, then the value of $$n$$ is:

A
$$12$$

B
$$11$$

C
$$10$$

D
$$9$$

Solution
Question 10
1 / -0

Total number of ways of selecting two numbers from the set $${1,2,3,...90}$$ so that their sum is divisible by $$3$$ is

A
$$885$$

B
$$1335$$

C
$$1770$$

D
$$3670$$

Solution