 Crampete

#### 1)In how many ways Raju can distribute chocolates for his 6 friends on his birthday if he has 4 chocolates?

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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).

#### 2)In how many ways can the letters of the word “MANAGEMENT” be arranged?

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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).

#### 3)How many distinct 5 digit numbers can be formed from the digits 2, 5, 6, 7, 8 and 9 whicharedivisible by 5?

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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).

#### 4)18 points lie on a circle. How many cyclic triangles can be drawn by using these points?

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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).

#### 5)There are 15 Men and 8 Women in a team. In how many ways can 3 Men and 2 Women be selected as Managers?

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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).

#### 6)How many words can be formed from the letters of word “PRELIMINARY” such that the vowelsalways come together?

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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).

#### 7)17 people at a party shake hands once with every other person in the party hall. How many handshakes took place?

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There are  17 people handshaking with every other person.
Two people will  constitute one handshake, 17c 2 =  $${17 * 16 \over { 2}}$$ = 136 Hence (b).

#### 8)How many different necklaces can be formed with 5 Pink beads and 6 Blue beads?

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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).

#### 9)How many 5 letter words can be formed using English alphabets such that last two letters is always a consonant (without repetition)?

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Number of ways to form 5 letter words =  24 23222120

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

#### 10)Find the rank of the word ‘SPEND’.

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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 .

D421 = 24

E421 = 24
N421 = 24

P421 = 24

SD21 = 6

SE21 = 6

SN21 = 6

SPD21 = 2

SPED1 = 1

SPEND = 1

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

#### 11)A committee is to be formed comprising8 members such that there is a simple majority of girlsand at leastoneboy. The shortlist consists of10 girls and9 boys. In how many ways can thiscommittee be formed?

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Total number of ways = (9c 1 * 10c 7 ) + (9c 2 * 10c 6 ) + (9c 3 * 10c 5 )
= 1080 + 7560 + 252
= 8892.  Hence (d).

#### 12)Find the sum of all numbers that can be formed with the digits 3, 2, 1 and 5 taken all at a time.

Show Solution
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).

#### 13)Find the rank of the word ‘ERODE’.

Show Solution
The alphabetical order of ERODE is DEEOR.

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

421 =  $${4! \over { 2!}}$$ = 12

D21 =  $${3! \over { 2!}}$$ = 3
EE21 =  $${3! \over { 2!}}$$ = 3

EO21 =  $${3! \over { 2!}}$$ = 3

ERD21 =  $${2! \over { 2!}}$$ = 1

ERE21 =  $${2! \over { 2!}}$$ = 1

ERODE = 1

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

#### 14)How many factors of27* 54 * 119 are perfect squares?

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Any factor of this number should be of the form 2 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).

#### 15)When five fair dice are rolled simultaneously, in how many outcomes will at least one of the dice show 2?

Show Solution
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).