#### Permutation And Combination

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

^{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?

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 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).

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

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 vowels

always come together?

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?

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?

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)?

__24__*

__23__*

__22__*

__21__*

__20__

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

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

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).

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

_{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.

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’.

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).

**14)** How many factors of2^{7}* 5^{4} * 11^{9} are perfect squares?

^{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).

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

^{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).