Permutations and Combinations Test 39

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Permutations and Combinations Test 39

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Weekly Quiz Competition

Question 1
1 / -0

The total number of ways of arranging the letters $$AAABBBCCDEF$$ in a row such that letters $$C$$ are separated from one another is

A
$$277200$$

B
$$138600$$

C
$$453600$$

D
none of these

Solution
Question 2
1 / -0

If $$64a^2+36b^2=400, ab=4 $$ , then $$8a+6b$$ is:

A
26

B
28

C
22

D
None of the above

Solution

Given $$64a^2+36b^2=400,\quad ab=4$$
We know
$$(a+b)^2=a^2+b^2+2ab$$
therefore,
$$(8a+6b)^2=(8a)^2+(6b)^2+2(8a)(6a)$$
$$=64a^2+36b^2+96ab$$
$$=400+96\times4$$
$$=400+384$$
$$(8a+6b)^2=784$$
$$8a+6b=\sqrt{784}$$
$$8a+6b=28$$
Question 3
1 / -0

There is a defined relationship between the pair of figure on either side of : :. Identify the relationship of the given pair and find be the missing figure.

A

B

C

D

Solution
Question 4
1 / -0

The number of positive integral solutions of the equation $$x _ { 1 } x _ { 2 } x _ { 3 } x _ { 4 } x _ { 5 } = 1050$$ is

A
1800

B
1600

C
1400

D
1875

Solution

We have,
$${x_1}{x_2}{x_3}{x_4}{x_5} = 1050 = 2 \times 3 \times {5^2} \times 7$$
Now,
$$2,3$$ and $$7$$ can be put in any boxes $${x_1},{x_2},{x_3},{x_4}$$ and $${x_5}$$.
Also $$5,5$$ can be distributed in $$5$$ boxes in
$$^{2 + 5 - 1}{C_{5 - 1}}{ = ^6}{C_4} = 15$$Ways.
So total no. of positive integral solutions $$=15 \times5 \times5 \times5$$
$$=1875$$
Option $$D$$ is correct answer.
Question 5
1 / -0

If the area of triangle formed by the points $$(2a,b),(a+b,2b+a)$$ and $$(2b,2a)$$ be $$\lambda$$ then the area of the triangle whose vertices are $$(a+b,a-b), (3b-a,b+3a)$$ and $$(3a-b,3b-a)$$ will be

A
$$\dfrac{3}{2} \lambda$$

B
$$3\lambda$$

C
$$4\lambda$$

D
$$none\ of\ these$$

Solution

Area of $$ \triangle \,= \dfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right |$$
Given area of triangle whose vertices are $$ (2a,b), (a+b,2b+a), (2b,2a)$$ is $$ \lambda $$
$$\therefore \lambda = \dfrac{1}{2}\left | 2a(2b+a-2a)+(a+b)(2a-b)+2b(b-2b-a) \right | $$
$$ \lambda = \dfrac{1}{2}\left | 4ab+2a^{2}-4a^{2}+2a^{2}-ab+2ab-b^{2}+2b^{2}-4b^{2}-2ab \right |$$
$$ \lambda = \dfrac{1}{2}\left | 3ab - 3b^{2} \right | $$ __ (1)
Now area of triangle whose vertices are $$ (a+b,a-b), (3b-a,b+3a), (3a-b,3b-a)$$ is given by
Area $$ = \dfrac{1}{2}\left | (a+b)(b+3a-3b+a)+(3b-a)(3b-a-a+b)+(3a-b)(a-b-b-3a) \right |$$
$$ = \dfrac{1}{2}\left | (a+b) (-2b+4a)+(3b-a)(4b-2a)+(3a-b)(-2a-2b) \right |$$
$$ = \dfrac{1}{2}\left | -2ab+4a^{2}-2b^{2}+4ab+12b^{2}-6ab-4ab+2a^{2}-6a^{2}-6ab+2ab+2b^{2} \right |$$
$$ = \dfrac{1}{2}\left | 12b^{2}-12ab \right |$$
$$ = \dfrac{4}{2}\left | -(3ab-3b^{2}) \right |$$
$$ = 4.\dfrac{1}{2}\left | 3ab-3b^{2} \right |$$
$$ = 4 \lambda .$$
Question 6
1 / -0

If $$(1+x+x^2)^n=\displaystyle\sum^{2n}_{r=0}a_rx^r$$, then $$a_0a_{2r}-a_1a_{2r+1}+a_2a_{2r+2}-....=?$$

A
$$a_r$$

B
$$a_{n-2r}$$

C
$$a_{n+r}$$

D
$$a_{2r}$$
Question 7
1 / -0

If repetitions are not allowed, the number of numbers consisting of $$4$$ digits and divisible by $$5$$ and formed out of $$0,1,2,3,4,5,6$$ is

A
$$220$$

B
$$240$$

C
$$370$$

D
$$588$$

Solution

$$\rightarrow$$ For divisibility by $$5$$, unit's place should be $$0$$ or $$5$$
with $$0$$, ----$$0$$
$$6$$ choices $$\Rightarrow ^6C_3\times 3!$$
$$=\dfrac{6!}{3!3!}\times 3!=6.5.4$$
$$=120$$
with $$5$$, ----$$5$$
$$6$$ choices $$=120$$
$$^5C_2\times 2!=10\times 2=20$$
$$\Rightarrow 120+120-20=220$$
$$\rightarrow (A)$$
Question 8
1 / -0

A Printer number the pages of a book starting with $$1$$ and used $$3193$$ digit in all. How many pages does the book have:

A
$$1074$$

B
$$1120$$

C
$$1595$$

D
None of these

Solution

No. of digits in 1-digit page no. $$\Rightarrow 1\times 9=9$$
No. of digits in 2-digit page no. $$\Rightarrow 2\times 90= 180$$
No. of digits in 3-digit page no. $$\Rightarrow 3\times 900= 2700$$
No. of digits in 4-digit page no.$$ \Rightarrow 300$$
$$\therefore $$ No. of pages with 4- digit page no. $$= (300/4)=75$$
Hence, total no. of pages in the book $$= (999+75)= 1074$$
Question 9
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 10
1 / -0

$$\sum _{ k=1 }^{ 10 }{ k.k! } =$$

A
10!

B
11!

C
10!+1

D
11!-1

Solution

Given,

$$\sum _{k=1}^{10}kk!$$

$$a_k=kk!$$

$$a_1=1\cdot \:1!=1$$

$$a_2=2\cdot \:2!=4$$

Similarly,

$$a_3=18$$

$$a_4=96$$

$$a_5=600$$

$$a_6=4320$$

$$a_7=35280$$

$$a_8=322560$$

$$a_9=3265920$$

$$a_{10}=36288000$$

$$=1+4+18+96+600+4320+35280+322560+3265920+36288000$$

$$=39916799$$

$$=11!-1$$