Permutations and Combinations Test 38

Result

Permutations and Combinations Test 38

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Question 1
1 / -0

In the following question, groups of letters are given. Find the odd one out.

A
$$\mathrm {EfgiL}$$

B
$$\mathrm {RcjkY}$$

C
$$\mathrm {DacsO}$$

D
$$\mathrm {QuegP}$$

Solution
Question 2
1 / -0

Let the eleven letters, $$A, B, ....K$$ denote an artbitrary permutation of the integers $$(1,2,....11)$$, then $$(A-1)(B-2)(C-3)...(K-11)$$ is

A
Necessarily zero

B
Always odd

C
Always evem

D
None of these

Solution

Given set of numbers is {1,2,3.....11} in which 5 are even and six are odd, which demands that in the given product it is not possible to arrange to subtract only even number from odd numbers. There must be at least one factor involving subtraction of an odd number from another odd number. So at least one of the factors is even. Hence product is always even.
Question 3
1 / -0

The maximum number of points of intersection of $$7$$ staright lines and $$5$$ circles when $$3$$ straight lines are parallel and $$2$$ circles are concentric,is /are

A
$$106$$

B
$$96$$

C
$$90$$

D
$$None\ of\ these$$

Solution
Question 4
1 / -0

Choose the odd one out .

A
KWIJ

B
TPRN

C
CHAF

D
OZMX

Solution
Question 5
1 / -0

Consider all permutations of the letters of the word MORADABAD.
The number of permutations which contain the word BAD is:

A
$$21 \times 5!$$

B
$$7 \times 5!$$

C
$$6 \times 5!$$

D
$$2 \times 5!$$

Solution

Total number of letters = $$9$$

Consider BAD as a group, then the number of total letters = $$6+1=7$$

$$\therefore$$ Permutation of 7 things = $$7!$$

but letters remaining after removing BAD will have two A's

$$\therefore$$ Total permutation = $$\dfrac{7!}{2}$$ = $$21\cdot 5!$$
Question 6
1 / -0

If $$x$$ and $$y$$ are the number of possibilities that $$A$$ can assume such that the unit digit of A and $$A^3$$ are same and the unit digit of $$A^2$$ and $$A^3$$ are same respectively ,then the value of $$x-y$$ is (where $$A$$ is a single digit number)

A
$$4$$

B
$$2$$

C
$$3$$

D
$$5$$

Solution

Here, according to the question
if $$A=1$$, then $${ A }^{ 3 }={ 1 }$$,
We see that unit digit number of $$A$$ and $${ A }^{ 3 }$$ are same.
for, $$A=2\Rightarrow { A }^{ 3 }=8,$$, unit digits are not same
$$A=3\Rightarrow { A }^{ 3 }=27,$$ unit digits are not same
$$A=4\Rightarrow { A }^{ 3 }=64,$$ unit digits are same
$$A=5\Rightarrow { A }^{ 3 }=125,$$ unit digits are same
$$A=6\Rightarrow { A }^{ 3 }=216,$$ unit digits are not same
$$A=7\Rightarrow { A }^{ 3 }=343,$$ unit digits are not same
$$A=8\Rightarrow { A }^{ 3 }=512,$$ unit digits are not same
$$A=8\Rightarrow { A }^{ 3 }=512,$$ unit digits are same
So, there are five possible solutions where unit digits of $$A$$ and $${ A }^{ 3 }$$ are same.
$$\therefore x=5$$
Again, if we take $$A=1,{ A }^{ 2 }=1,{ A }^{ 3 }={ 1 }$$, unit digits are same
$$A=2\Rightarrow { A }^{ 2 }=4,{ A }^{ 3 }=8,$$ unit digits are not same
$$A=3\Rightarrow { A }^{ 2 }=9,{ A }^{ 3 }=27,$$ unit digits are not same
$$A=4\Rightarrow { A }^{ 2 }=16,{ A }^{ 3 }=64,$$ unit digits are not same
$$A=5\Rightarrow { A }^{ 2 }=25,{ A }^{ 3 }=125,$$ unit digits are same
$$A=6\Rightarrow { A }^{ 2 }=36,{ A }^{ 3 }=216,$$ unit digits are same
$$A=7\Rightarrow { A }^{ 2 }=49,{ A }^{ 3 }=343,$$ unit digits are not same
$$A=8\Rightarrow { A }^{ 2 }=64,{ A }^{ 3 }=512,$$ unit digits are not same
$$A=9\Rightarrow { A }^{ 2 }=81,{ A }^{ 3 }=729$$ unit digits are not same
Here, we can see that there are three possible solutions where unit digits of $${ A }^{ 2 },{ A }^{ 3 }$$ are same.
So, $$y=3$$
Hence, $$x-y=5-3=2.$$
Question 7
1 / -0

The number of permutation of the letters of the word $$HINDUSTAN$$ such that neither the pattern $$'HIN'$$ nor $$'DUS'$$ nor $$'TAN'$$ appears, are :

A
$$166674$$

B
$$169194$$

C
$$166680$$

D
$$181434$$

Solution

Total number of letters $$=9$$ in which N is repeated twice

Therefore total number of permutation $$=\dfrac{9!}{2}$$

The number of permutation in which HIN comes as block $$=9-3+1=7!$$

The number of permutation in which DUS comes as block $$=7!$$

The number of permutation in which TAN comes as block $$=\dfrac{7!}{2}$$

The number of permutation in which HIN and DUS comes as block $$=5!$$

The number of permutation in which DUS and TAN comes as block $$=5!$$

The number of permutation in which HIN and TAN comes as block $$=5!$$

Therefore the required number of permutations $$=\dfrac{9!}{2}-\left(7!+7!+\dfrac{7!}{2}-3\times5!+3!\right)$$

$$=\dfrac{362880}{2}-\left(5040+5040+\dfrac{5040}{2}-360+6\right)$$

$$=181440-\left(10080+2520-360+6\right)$$

$$=181440-12246$$

$$=169194$$
Question 8
1 / -0

In how many ways atleast one horse and atleast one dog can be selected out of eight horses and seven dogs.

A
$${2}^{15}-2$$

B
$${2}^{15}-1$$

C
$$({2}^{8}-1)({2}^{7}-1)$$

D
$${ _{ }^{ 15 }{ C } }_{ 2 }$$

Solution

As we can either select or not select any horse; total ways for horses$$=2^r$$ ways to be excluded are when no horse is selected, which is only one way, therefore, favourable ways for horses$$=2^8-1$$
similarly, favourable ways for dogs$$=2^7-1$$
$$\Rightarrow$$ Total ways$$=(2^8-1)(2^7-1)$$.
Question 9
1 / -0

$$\begin{bmatrix} 12 & 47 & 21 \\ 10 & 52 & 4 \\ 64 & ? & 24 \end{bmatrix}$$

A
$$40$$

B
$$83$$

C
$$62$$

D
$$16$$

Solution

$$\begin{bmatrix} 12 & 47 & 21\\ 10 & 52 & 4\\ 64 & ? & 24\end{bmatrix}$$
We can see that dividing third number in each row with second digit of middle number and multiplying with first digit gives first member.
Similarly, out of $$4$$ options, dividing $$24$$ by $$3$$ and multiplying by $$8$$ gives $$64$$.
$$\Rightarrow$$ Option $$(B)$$.
Question 10
1 / -0

The number of ways in which the letters of the word $$"ARRANGE"$$ can be permuted such that $$R's$$ occur together is

A
$$\dfrac{!7}{!2!2}$$

B
$$6!$$

C
$$\dfrac{6!}{2!}$$

D
none of these

Solution