# Permutation Combination Practice test

A box contains 3 white balls, 4 black balls and 5 yellow balls. In how many ways can 4 balls be drawn from the box, if at least one yellow ball is to be included in the draw?

**Answer:**Option C

**Explanation:**

We may have (1 yellow and 3 others) or (2 yellow and 2 others) or (3 yellow and 1 others) or (4 yellow).

Therefore, Required number of ways= (^{4}C_{1}*^{8}C_{3})+ (^{4}C_{2}*^{8}C_{2})+ (^{4}C_{3}*^{8}C_{1})+ (^{4}C_{4})

= 4* 8*7*6 + 4*3 * 8*7 + (^{4}C_{1}*8) +1

3*2*1 2*1 2*1

= 224+168+32+1= 425

There are 6 red shoes & 4 green shoes . If two of red shoes are drawn what is the probability of getting red shoes

**Answer:**Option A

**Explanation:**

Total no.of shoes=6+4=10;

from total two shoes drawn=^{10}c_{2}

two of red shoes are drawn=^{6}c_{2}

so from 10 shoes propability of drawing two red shoes=^{6}c_{2}/^{10}c_{2}

6 parallerl lines are intersected by 5 other parallel lines. The total number of parallelogram formed is

**Answer:**Option B

**Explanation:**

Any two lines in one direction and any two parallel lines in the other direction can form a parallograms.

So, number of parallogram formed

= 6c2 x 5c2 = 6!/(4!*2!) +5!/(3!*2!) =150

In how many ways can 21 books on English and 19 books on Hindi be placed in a row on a shelf so that two books on Hindi may not be together?

**Answer:**Option B

**Explanation:**

In order that two books on Hindi are never together, we must place all these books as under:

X E X E X E X …. X E X

Where E – denotes the position of an English book and X that of a Hindi book.

Since there are 21 books on English, the number of places marked X are therefore, 22. Now, 19 places out of 22 can be chosen in

22C_{19} = ^{22}C_{3} = (22 x 21 x 20)/3 x 2 x 1 = 1540 ways

Hence, the required number of ways = 1540.

How many words can be formed by re-arranging the letters of the word ASCENT such that A and T occupy the first and last position respectively?

**Answer:**Option B

**Explanation:**

As A and T should occupy the first and last position, the first and last position can be filled in only one way. The remaining 4 positions can be filled in 4! Ways by the remaining words (S,C,E,N,T). hence by rearranging the letters of the word ASCENT we can form 1×4! = 4! words.

Find the number of ways in which one or more letter be selected from the letters AAAABBCCCDEF.

**Answer:**Option A

**Explanation:**

The number of ways of

Selecting A = 4+1 = 5

Selecting B = 2+1 = 3

Selecting C = 3+1 = 4

Selecting D = 1+1 = 2

Selecting E = 1+1 = 2

Selecting F = 1+1 = 2

Hence, total ways of selecting =

How many permutations of 3 different digits are there, chosen from the ten digits 0 to 9 inclusive?

**Answer:**Option C

**Explanation:**

The number of permutations of 3 digits chosen from 10 is ^{10}P_{3} = 10 × 9 × 8 = 720

It is required to seat 7 men and 3 women in a row such that women should occupy even places how many such arrangements are possible ?

**Answer:**Option B

**Explanation:**

All the 10 persons should be arranged in a row

And in a row of 10 positions there are exactly 5 even places.

3 of these even places should be occupied by 3 women

this can be done in Ways.

The remaining 7 positions can be filled by the 7 men in ^{7} p _{7}Ways

Hence, the total number of seating arrangements

= ^{5}P _{3} x ^{7} P _{7}

= 302400

How many different four letter words can be formed (the words need not be meaningful) using the letters of the word “MEDITERRANEAN” such that the first letter is E and the last letter is R?

**Answer:**Option C

**Explanation:**

The first letter is E and the last one is R.

Therefore, one has to find two more letters from the remaining 11 letters.

Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters.

The second and third positions can either have two different letters or have both the letters to be the same.

**Case 1**: When the two letters are different. One has to choose two different letters from the 8 available different choices. This can be done in 8 * 7 = 56 ways.

**Case 2**: When the two letters are same. There are 3 options – the three can be either Ns or Es or As. Therefore, 3 ways.

Total number of posssibilities = 56 + 3 = 59.

Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

**Answer:**Option D

**Explanation:**

Here, S = {1, 2, 3, 4, …., 19, 20}.

Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

P(E)= n(E)/n(S)= 9/20