Permutations and Combinations Test 1

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Permutations and Combinations Test 1

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Weekly Quiz Competition

Question 1
1 / -0

All possible number are formed using the digits $$1,1,2,2,2,2,3,4,4$$ taken all at a time. The number of such number in which the odd digits occupy even places is:

A
$$175$$

B
$$162$$

C
$$160$$

D
$$180$$

Solution

Number of such number $$=^4C_3\times \dfrac{3!}{2!}\times \dfrac{6!}{2!4!}=180$$
Question 2
1 / -0

The value of $$(^{21}C_1-^{10}C_1)+(^{21}C_2-^{10}C_2)+(^{21}C_3-^{10}C_3)+(^{21}C_4+^{10}C_4)+......+(^{21}C_{10}-^{10}C_{10})$$ is:

A
$$2^{21}-2^{11}$$

B
$$2^{21}-2^{10}$$

C
$$2^{20}-2^9$$

D
$$2^{20}-2^{10}$$

Solution

$$(^{21}C_1-^{10}C_1)+(^{21}C_2-^{10}C_2)+(^{21}C_3-^{10}C_3)+(^{21}C_4+^{10}C_4)+......+(^{21}C_{10}-^{10}C_{10})$$

$$= ^{21}C_1+^{21}C_2+^{21}C_3+ ..... + ^{21}C_{10} - [2^{10}-1]$$

$$=\cfrac 12 [2 ^{21}C_1 + 2 ^{21}C_2+ 2 ^{21}C_3 + ..... + 2 ^{21}C_{10}] - [2^{10}-1]$$

$$=\cfrac 12 [^{21}C_1 + ^{21}C_2+ .... + ^{21}C_{10} + ^{21}C_{11} + ..... + ^{21}C_{19} + ^{21}C_{20}] - [2^{10}-1]$$

$$=\cfrac 12 [2^{21} - 1 -1 ] - [2^{10}-1]$$

$$=2^{20} - 1 - 2^{10} +1$$

$$=2^{20}-2^{10}$$
Question 3
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A group of students comprises of $$5$$ boys and $$n$$ girls. If the number of ways, in which a team of $$3$$ students can randomly be selected from this group such that there is at least one boy and at least one girl in each team, is $$1750$$, then $$n$$ is equal to

A
$$25$$

B
$$28$$

C
$$27$$

D
$$24$$

Solution

Number of ways of selecting a team with 1 boy and 2 girls$$=^{5}C_{1}. ^{n}C_{2} $$

Number of ways of selecting a team with 2 boy and 1 girls$$=^{5}C_{2} . ^{n}C_{1} $$

Given $$^{5}C_{1}. ^{n}C_{2} + ^{5}C_{2} . ^{n}C_{1} = 1750$$

$$5\cdot \dfrac {n(n-1)}{2!}+10\cdot n=1750$$

$$n^{2} + 3n = 700$$

$$\therefore n = 25$$.
Question 4
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The number of four -digit numbers strictly grater than $$4321$$ that can be formed using the digits $$0,1,2,3,4,5$$ (repetition of digits is allowed) is:

A
$$288$$

B
$$306$$

C
$$360$$

D
$$310$$

Solution

(1) The number of four-digit numbers Strating with 5 is equal to $$6^3 = 216$$
(2) Starting with 44 and 55 is equal to $$36\times 2 = 72$$
(3) Starting with 433, 434 and 435 is equal to $$6\times 3 = 18$$
(4) Remaining numbers are 4322, 4323, 4324, 4325 is equal to 4
so total number are
$$716 + 72 + 18 + 4 = 310$$
Question 5
1 / -0

Let $$A$$ be the set of all $$3 \times 3$$ symmetric matrices all of whose entries are either $$0$$ or $$1$$. Five of these entries are $$1$$ and four of them are $$0$$.
The number of matrices in $$A$$ is

A
$$12$$

B
$$6$$

C
$$9$$

D
$$3$$

Solution

If two zeros are the entries in the diagonal, then
$$^{3}\mathrm{C}_{2}\times^{3}\mathrm{C}_{1}$$
If all the entries in the principle diagonal is 1, then
$$^{3}\mathrm{C}_{1}$$
$$\Rightarrow$$ Total matrix $$= 12.$$
Question 6
1 / -0

In how many ways can $$4$$ people be seated on a square table, one on each side?

A
$$4!$$

B
$$3!$$

C
$$1$$

D
None of these

Solution

The number of circular arrangement for $$n$$ objects is $$(n-1)!.$$
Similar to arrangements in a circle, there would be $$3!$$ ways possible of making $$4$$ people sit on a square table.
Question 7
1 / -0

The number of ways four boys can be seated around a round table in four chairs of different colours is:

A
$$24$$

B
$$12$$

C
$$23$$

D
$$64$$

Solution

Since the chair are of different colours, so you can treat it as linear permutation
So number of ways will be $$4!=24$$
Question 8
1 / -0

Seven people are seated in a circle. How many relative arrangements are possible?

A
$$7!$$

B
$$6!$$

C
$$^7P_6$$

D
$$^7C_6$$

Solution

The number of circular arrangement for $$n$$ objects is $$(n-1)!.$$
There would be $$(7 - 1)! = 6!$$ arrangements possible.
Question 9
1 / -0

Value of 0! is always 1.

A
True

B
False

C
Either

D
Neither

Solution

we know $$1!=1$$
Also
$$n!=n\times (n-1)\times (n-2)........3\times 2\times 1\\ n!=n\times (n-1)!\\ 1!=1(1-1)!\\ 1=1(0)!\\ 0!=1$$
$$0! $$ is always $$1$$
Hence its true that $$0!$$ is always $$1$$
Question 10
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Each combination corresponds to many permutations.

A
True

B
False

C
Either

D
Neither

Solution

In combination
Each combination can be considered as a set of selection an order
Each selection has a defined order
They can be considered as a permutation
Each cpmbination corresponds to many permutations
Hence the above statement is true.

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Re-Attempt Weekly Quiz Competition

Permutations and Combinations Test 1

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Permutations and Combinations Test 1

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SHARING IS CARING

Question 1

$$175$$

$$162$$

$$160$$

$$180$$

Solution

Question 2

$$2^{21}-2^{11}$$

$$2^{21}-2^{10}$$

$$2^{20}-2^9$$

$$2^{20}-2^{10}$$

Solution

Question 3

$$25$$

$$28$$

$$27$$

$$24$$

Solution

Question 4

$$288$$

$$306$$

$$360$$

$$310$$

Solution

Question 5

$$12$$

$$6$$

$$9$$

$$3$$

Solution

Question 6

$$4!$$

$$3!$$

$$1$$

None of these

Solution

Question 7

$$24$$

$$12$$

$$23$$

$$64$$

Solution

Question 8

$$7!$$

$$6!$$

$$^7P_6$$

$$^7C_6$$

Solution

Question 9

True

False

Either

Neither

Solution

Question 10

True

False

Either

Neither

Solution

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