Permutations and Combinations Test 41

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Permutations and Combinations Test 41

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Weekly Quiz Competition

Question 1
1 / -0

$$14X's$$ have to be placed in the squares of the above figure such that each row contains at least one $$X$$. In how many ways can this be done?

A
96

B
104

C
112

D
118

Solution

Total X's =14
Total Blocks = 16
2 extra blocks,
only possible way is if no X is put either in the first row or the last row
No. of ways = Total ways of Putting X's - 2
= $$^{16}C_{14} -2$$
=$$118$$
Question 2
1 / -0

If all the permutations of the letters in the word 'OBJECT' are arranged (and numbered serially) in alphabetical order as in a dictionary, then the $$717^{th}$$ word is

A
TOJECB

B
TOEJBC

C
TOCJEB

D
TOJCBE

Solution

The order of letters of the word "OBJECT' is B C E J O T
Words starting with:
B can be formed in $$5!=120$$
C can be formed in $$5!=120$$.Total = 240
E can be formed in $$5!=120$$. Total = 360
J can be formed in $$5!=120$$. Total = 480
O can be formed in $$5!=120$$. Total = 600
TB can be formed in $$4!=24$$. Total = 624
TC can be formed in $$4!=24$$. Total = 648
TE can be formed in $$4!=24$$. Total = 672
TJ can be formed in $$4!=24$$. Total = 696
TO can be formed in $$4!=24$$. Total = 720
So, we have exceeded 717. We need to go back by 3.
Last word with TO is TOJECB ---- number 720
Before that comes TOJEBC ----- number 719
Before that comes TOJCEB ----- number 718
Before that comes TOJCBE ----- number 717
Hence, (D) is correct.
Question 3
1 / -0

The value of $$^{47}C_4 + \displaystyle\sum_{r=1}^{5}{^{52-r}C_3}$$=

A
$$^{52}C_2$$

B
$$^{52}C_3$$

C
$$^{52}C_4$$

D
$$^{52}C_5$$

Solution

We know that,
$$ \displaystyle ^{n}{C}_{r}+{}^{n}{C}_{r-1}$$

$$\displaystyle =\frac{n!}{r!(n-r)!}+\frac{n!}{(r-1)!(n-r+1)!}$$

$$\displaystyle =\frac{n!(n-r+1)}{r!(n-r)!(n-r+1)}+\frac{n!r}{(r-1)!(n-r+1)!r}$$

$$\displaystyle =\frac{n!(n-r+1)}{r!(n-r+1)!}+\frac{n!r}{(n-r+1)!r!}$$

$$\displaystyle =\frac{n!}{r!(n-r+1)!}(n-r+1+r)$$

$$\displaystyle =\frac{n!}{r!(n-r+1)!}(n+1)$$

$$\displaystyle =\frac{(n+1)!}{r!(n-r+1)!}$$

$$=\displaystyle {}^{n+1}{C}_{r}$$

$$\Rightarrow \displaystyle {}^{n}{C}_{r}+{}^{n}{C}_{r-1}={}^{n+1}{C}_{r}$$
The equation is: $$({}^{47}{C}_{4}+{}^{47}{C}_{3})+{}^{48}{C}_{3}+{}^{49}{C}_{3}+{}^{50}{C}_{3}+{}^{51}{C}_{3}$$
$$=({}^{48}{C}_{4}+{}^{48}{C}_{3})+{}^{49}{C}_{3}+{}^{50}{C}_{3}+{}^{51}{C}_{3}$$
$$=({}^{49}{C}_{4}+{}^{49}{C}_{3})+{}^{50}{C}_{3}+{}^{51}{C}_{3}$$
$$=({}^{50}{C}_{4}+{}^{50}{C}_{3})+{}^{51}{C}_{3}$$
$$=({}^{51}{C}_{4}+{}^{51}{C}_{3})$$
$$={}^{52}{C}_{4}$$
Hence, option 'C' is correct.
Question 4
1 / -0

If the difference of the number of arrangements of three things from a certain number of dissimilar things and the number of selections of the same number of things from them exceeds $$ 100$$, then the least number of dissimilar things is

A
$$8$$

B
$$6$$

C
$$5$$

D
$$7$$

Solution

Here,
$$^nP_3-^nC_3 > 100$$

or $$\dfrac {n!}{(n-3)!}-\dfrac {n!}{3!(n-3)!} > 100$$

or $$\dfrac {5}{6}n(n-1)(n-2) > 100$$

or $$n(n-1)(n-2) > 100$$

or $$n(n-1)(n-2) > 6\times 5\times 4$$

or $$n=7, 8, .....$$
Question 5
1 / -0

$$p$$ is a prime number and $$n < p < 2n$$. If $$N=^{2n}C_n$$, then

A
$$p$$ divides $$N$$ completely

B
$$p^2$$ divides $$N$$ completely

C
$$p$$ cannot divide $$N$$

D
none of these

Solution

Solution:
$$N={}^{2n}C_n=\cfrac{2n!}{n!n!}=\cfrac{(n+1)(n+2)(n+3)........(2n)}{(1).(2).......(n)}$$
As $$p>n$$ and $$p<2n, p$$ occurs exactly once in the numerator but doesn't occur in the denominator.
It means $$p$$ divides $$N$$ completely.
Hence, A is the correct option.
Question 6
1 / -0

The rank of the word $$NUMBER$$ obtained, if the letters of the word $$NUMBER$$ are written in all possible orders and these words are written out as in a dictionary is

A
$$468$$

B
$$469$$

C
$$470$$

D
$$471$$

Solution

Alphabetically,
Starting with B $$5!$$ ways
Starting with E $$5!$$ ways
Starting with M $$5!$$ ways
Starting with NB $$4!$$ ways
Starting with NE $$4!$$ ways
Starting with NM $$4!$$ ways
Starting with NU $$4!$$ ways
Starting with NUB $$3!$$ ways
Starting with NUE $$3!$$ ways
Starting with NUMBER $$1!$$ ways
Therefore, the total will be $$3.5!+4.4!+2.3!+1=469$$ ways
Hence, option 'B' is correct.
Question 7
1 / -0

$$^nC_r + \displaystyle\sum_{j=0}^{3}{^{n+j}C_{r+1+j}} = $$_________________

A
$$^{n+3}C_{r+3}$$

B
$$^{n}C_{r+4}$$

C
$$^{n+1}C_{r+4}$$

D
$$^{n+4}C_{r+4}$$

Solution

Expanding
We get
$$\:^{n}C_{r}+\:^{n}C_{r+1}+\:^{n+1}C_{r+2}+\:^{n+2}C_{r+3}+\:^{n+3}C_{r+4}$$
$$=\:^{n+1}C_{r+1}+\:^{n+1}C_{r+2}+\:^{n+2}C_{r+3}+\:^{n+3}C_{r+4}$$
$$=\:^{n+2}C_{r+2}+\:^{n+2}C_{r+3}+\:^{n+3}C_{r+4}$$
$$=\:^{n+3}C_{r+3}+\:^{n+3}C_{r+4}$$
$$=\:^{n+4}C_{r+4}$$
Question 8
1 / -0

The number of factors (excluding $$1$$ and the expression itself) of the product of $$a^7b^4c^3def$$ where $$a,b,c,d,e,f$$ are all prime numbers is

A
$$1278$$

B
$$1360$$

C
$$1100$$

D
$$1005$$

Solution

A factor of expression $$a^{7}b^{4}c^{3}def$$ is simply the result of selecting one or more letters from $$7 a's, 4 b's, 3a's, d, e, f.$$
The collection of letters can be observed as a collection of $$17$$ objects out of which $$7$$ are alike of one kind (a's), $$4$$ are of second kind (b's), $$3$$ are of third kind (c' s) and $$3$$ are different $$(d, e, f)$$
The number of selections =$$ (1 + 7) (1 + 4) (1 + 3)(1+1)^{3} = 8 \times5\times 4\times 8 = 1280. $$
But we have to exclude two cases : (i) When no letter is selected (ii) When all are selected.
So, the number of factors $$= 1280 -2 = 1278$$
Hence, option $$'A'$$ is correct.
Question 9
1 / -0

If $$\space ^{n+1}C_{r+1}: ^nC_r : ^{n-1}C_{r-1} = 11:6:3$$, then find the values of $$n$$ and $$r$$.

A
$$n=10,\space r=5$$

B
$$n=9,\space r=4$$

C
$$n=11,\space r=5$$

D
$$n=10,\space r=4$$

Solution

Simplifying , we get
$$6\:^{n+1}C_{r+1}=11\:^{n}C_{r}$$
$$\Rightarrow 6(n+1)=11(r+1)$$
$$\Rightarrow 6n-11r=5$$ ... (i)
Also
$$3\:^{n}C_{r}=6\:^{n-1}C_{r-1}$$
$$\Rightarrow 3n=6r$$ ...(ii)
Substituting in 1, we get
$$12r-11r=5$$
$$\Rightarrow r=5$$
Hence
$$n=10$$
Question 10
1 / -0

There are three papers of $$100$$ marks each in an examination. In how many ways can a student get $$150$$ marks such that he gets at least $$60%$$ in two papers?

A
$$1480$$

B
$$1488$$

C
$$1520$$

D
$$1526$$

Solution

Suppose the student gets at least $$ 60%$$ marks in first two papers, then he just gets at most $$30$$ marks in the third paper to make a total of $$150$$ marks

So, the required numbers of ways = coefficient of $${ x }^{ 150 }$$ in $${ \left( { x }^{ 60 }+{ x }^{ 61 }+...+{ x }^{ 100 } \right) }^{ 2 }\left( 1+x+{ x }^{ 2 }+...+{ x }^{ 30 } \right) $$

$$=$$ coefficient of $${ x }^{ 30 }$$ in $$\left( { \left( 1+x+...+{ { x }^{ 40 } } \right) }^{ 2 }\left( 1+x+...{ x }^{ 30 } \right) \right) $$

$$=$$ coefficient of $${ x }^{ 30 }$$ in $$\displaystyle { \left( \frac { 1-{ x }^{ 41 } }{ 1-x } \right) }^{ 2 }\left( \frac { 1-{ x }^{ 31 } }{ 1-x } \right) $$

$$=$$ coefficient of $${ x }^{ 30 }$$ in $${ \left( 1-x \right) }^{ -3 }$$

$$=^{ 30+3-1 }{ C }_{ 3-1 }=^{ 32 }{ C }_{ 2 }$$

Thus the students get at least $$60%$$ marks in the first two papers to get 150 marks as total in $$^{ 32 }{ C }_{ 2 }$$ ways. But the two papers at least $$60%$$, can be chosen out of 3 papers in $$^{ 3 }{ C }_{ 2 }$$ ways.

Hence,the required number of ways $$^{ 3 }{ C }_{ 2 }\times ^{ 32 }{ C }_{ 2 }=1488$$

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Re-Attempt Weekly Quiz Competition

Permutations and Combinations Test 41

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Permutations and Combinations Test 41

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SHARING IS CARING

Question 1

96

104

112

118

Solution

Question 2

TOJECB

TOEJBC

TOCJEB

TOJCBE

Solution

Question 3

$$^{52}C_2$$

$$^{52}C_3$$

$$^{52}C_4$$

$$^{52}C_5$$

Solution

Question 4

$$8$$

$$6$$

$$5$$

$$7$$

Solution

Question 5

$$p$$ divides $$N$$ completely

$$p^2$$ divides $$N$$ completely

$$p$$ cannot divide $$N$$

none of these

Solution

Question 6

$$468$$

$$469$$

$$470$$

$$471$$

Solution

Question 7

$$^{n+3}C_{r+3}$$

$$^{n}C_{r+4}$$

$$^{n+1}C_{r+4}$$

$$^{n+4}C_{r+4}$$

Solution

Question 8

$$1278$$

$$1360$$

$$1100$$

$$1005$$

Solution

Question 9

$$n=10,\space r=5$$

$$n=9,\space r=4$$

$$n=11,\space r=5$$

$$n=10,\space r=4$$

Solution

Question 10

$$1480$$

$$1488$$

$$1520$$

$$1526$$

Solution

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