First friend has 4 options to receive the chocolate. It can be either1 ^{st} or 2 ^{nd} or 3 ^{rd} or 4 ^{th}

chocolate.

Similarly, the 2^{nd} , 3 ^{rd} , 4 ^{th} , 5 ^{th} and 6 ^{th} has 4 choice s each.

Each of the 6 friends has 4 choices to receive the chocolate

Required number of ways = 4 * 4 * 4 * 4 * 4 * 4 = 4096 Hence (a).

chocolate.

Similarly, the 2

Each of the 6 friends has 4 choices to receive the chocolate

Required number of ways = 4 * 4 * 4 * 4 * 4 * 4 = 4096 Hence (a).

The word MANAGEMENT has 10 letters that can be arrange d in 10! Ways. But in this word

letters M, A, N and E are repeated 2 times

Then, the numbers of ways of arranging the letters of the word MANAGEMENT are \({10! \over {2!*2!*2!*2!}}\)

Hence (b).

letters M, A, N and E are repeated 2 times

Then, the numbers of ways of arranging the letters of the word MANAGEMENT are \({10! \over {2!*2!*2!*2!}}\)

Hence (b).

How many distinct 5 digit numbers can be formed from the digits 2, 5, 6, 7, 8 and 9 which

aredivisible by 5?

For the numbers to be divisible by 5 last digit should be 0 or 5.

Here the unit digit à 5

Then, __5__* __4__ * __3 __* __2__ * __1__ = 120 numbers can be formed.Hence (d).

To form a cyclic triangle, we need 3 points to be selected out of 18 points.

Total number of ways of cyclic triangles that can be formed = 18c_{3} = \({18 * 17 * 16 \over {3 * 2}}\)

= 816 Hence (c).

Total number of ways of cyclic triangles that can be formed = 18c

= 816 Hence (c).

3 Men and 2 Women to be selected as Managers,

Number of ways = 15C_{3} * 8C _{2}

= \({15*14*13 \over {3*2}}\) * \({8*7 \over {2*1}}\) = 12740 Hence (b).

Number of ways = 15C

= \({15*14*13 \over {3*2}}\) * \({8*7 \over {2*1}}\) = 12740 Hence (b).

always come together?

In the word PRELIMINARY the vowels E, I, I and A always to be together, it can be treated as

One position.

Hence, the letters to be arranged are P, R, L, M, N, R, Y and (E, I, I, A).

There are 8 lett ers and it can be arranged in \({8! \over {2!}}\) Ways.

But the vowels can be arranged among themselves in \({4! \over {2!}}\) Ways.

Here the le tters I and R repeated.

Total number of ways = \({8!*4! \over {2!*2!}}\) Hence (c).

One position.

Hence, the letters to be arranged are P, R, L, M, N, R, Y and (E, I, I, A).

There are 8 lett ers and it can be arranged in \({8! \over {2!}}\) Ways.

But the vowels can be arranged among themselves in \({4! \over {2!}}\) Ways.

Here the le tters I and R repeated.

Total number of ways = \({8!*4! \over {2!*2!}}\) Hence (c).

There are 17 people handshaking with every other person.

Two people will constitute one handshake, 17c_{2} = \({17 * 16 \over { 2}}\) = 136 Hence (b).

Two people will constitute one handshake, 17c

Total beads = 11

Number of ways = \({10! \over {2!*5!*6!}}\) = 21

The arrangements are divided by 5! and 6!, as 5 Pink beads are alike and 6 Blue beads are alike.

Therefore, 21 necklaces can be formed. Hence (c).

Number of ways = \({10! \over {2!*5!*6!}}\) = 21

The arrangements are divided by 5! and 6!, as 5 Pink beads are alike and 6 Blue beads are alike.

Therefore, 21 necklaces can be formed. Hence (c).

Number of ways to form 5 letter words = __24 __ * __23__ * __22__ * __21__ * __20__

It can be in the form of \({24! \over {19!}}\) Hence (d).

It can be in the form of \({24! \over {19!}}\) Hence (d).

The alphabetical order of SPEND is DENPS .

If these words are arranged as in a dictionary, letter D will be will be in the first place .

__D__ * __4__ * __3 __ * __2__ * __1__ = 24

__E__ * __4__ * __3 __ * __2__ * __1__ = 24

__N__ * __4__ * __3 __ * __2__ * __1__ = 24

__P__ * __4__ * __3 __ * __2__ * __1__ = 24

__S__ * __D__ * __3 __ * __2__ * __1__ = 6

__S__ * __E__ * __3 __ * __2__ * __1__ = 6

__S__ * __N__ * __3 __ * __2__ * __1__ = 6

__S__ * __P__ * __D__ * __2__ * __1__ = 2

__S__ * __P__ * __E__ * __D__ * __1__ = 1

__S__ * __P__ * __E__ * __N__ * __D__ = 1

Rank of the word SPEND is ( 24 + 24 + 24 + 24 + 6 + 6+ 6 + 2 + 1 + 1) = 118 . Hence (a).

If these words are arranged as in a dictionary, letter D will be will be in the first place .

Rank of the word SPEND is ( 24 + 24 + 24 + 24 + 6 + 6+ 6 + 2 + 1 + 1) = 118 . Hence (a).

Total number of ways = (9c _{1} * 10c _{7} ) + (9c _{2} * 10c _{6} ) + (9c _{3} * 10c _{5} )

= 1080 + 7560 + 252

= 8892. Hence (d).

= 1080 + 7560 + 252

= 8892. Hence (d).

The 4 digits 3, 2, 1 and 5 can be arranged in 4! = 24 ways.

Out of 24, there are only 6 ways for each digits to occur in unit place.

Then, each digit will appear in 6 times.

The sum of numbers in unit’s place = (3+2+1+5)*6 = 66

The sum of numbers in ten’s place = (3+2+1+5)*6 = 66

The sum of numbers in hundred’s place = (3+2+1+5)*6 = 66

The sum of numbers in thousand’s place = (3+2+1+5)*6 = 66

Therefore, sum of all the four digit numbers = (66*1000) + (66*100) + (66*10) + (66*1)

= 73326 . Hence (c).

Out of 24, there are only 6 ways for each digits to occur in unit place.

Then, each digit will appear in 6 times.

The sum of numbers in unit’s place = (3+2+1+5)*6 = 66

The sum of numbers in ten’s place = (3+2+1+5)*6 = 66

The sum of numbers in hundred’s place = (3+2+1+5)*6 = 66

The sum of numbers in thousand’s place = (3+2+1+5)*6 = 66

Therefore, sum of all the four digit numbers = (66*1000) + (66*100) + (66*10) + (66*1)

= 73326 . Hence (c).

The alphabetical order of ERODE is DEEOR.

If these words are arranged as in dictionary, lette r D will be present first

__D __ * __4__ * __3 __ * __2__ * __1__ = \({4! \over { 2!}}\) = 12

__E __ * __D__ * __3 __ * __2__ * __1__ = \({3! \over { 2!}}\) = 3

__E__ * __E__ * __3 __ * __2__ * __1__ = \({3! \over { 2!}}\) = 3

__E__ * __O__ * __3 __ * __2__ * __1__ = \({3! \over { 2!}}\) = 3

__E__ * __R__ * __D__ * __2__ * __1__ = \({2! \over { 2!}}\) = 1

__E__ * __R__ * __E__ * __2__ * __1__ = \({2! \over { 2!}}\) = 1

__E__ * __R__ * __O__ * __D__ * __E__ = 1

Rank of the word ERODE is ( 12 + 3 + 3 + 3+ 1 + 1 + 1) = 24 . Hence (a).

If these words are arranged as in dictionary, lette r D will be present first

Rank of the word ERODE is ( 12 + 3 + 3 + 3+ 1 + 1 + 1) = 24 . Hence (a).

Any factor of this number should be of the form 2 ^{a } * 5 ^{b} * 11 ^{c}

For the factor to be a perfect square a , b , c have to be even.

a can take values 0 , 2 , 4, 6 . b can take values 0 , 2 . and c can take values 0 , 2, 4, 6 and 8 .

Total number of perfect squares = 4 * 2 * 5 = 40 . Hence (b).

For the factor to be a perfect square a , b , c have to be even.

a can take values 0 , 2 , 4, 6 . b can take values 0 , 2 . and c can take values 0 , 2, 4, 6 and 8 .

Total number of perfect squares = 4 * 2 * 5 = 40 . Hence (b).

When 5 dice are rolled simultaneously, there will be a total of 6 ^{5} = 7776 outcomes.

The number of outcomes in which none of the 5 dice show 2 will be 5^{5} = 3125 outcomes .

Therefore, the number of outcomes in whi ch at least one die will show 2 = 777 6 – 31 25

= 4651 . Hence (d).

The number of outcomes in which none of the 5 dice show 2 will be 5

Therefore, the number of outcomes in whi ch at least one die will show 2 = 777 6 – 31 25

= 4651 . Hence (d).