# Permutations and Combinations Questions

Q:

How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3 and 4, if repetition of digits is allowed?

 A) 376 B) 375 C) 500 D) 673

Explanation:

The smallest number in the series is 1000, a 4-digit number.

The largest number in the series is 4000, the only 4-digit number to start with 4.

The left most digit (thousands place) of each of the 4 digit numbers other than 4000 can take one of the 3 values 1 or 2 or 3.

The next 3 digits (hundreds, tens and units place) can take any of the 5 values 0 or 1 or 2 or 3 or 4.

Hence, there are 3 x 5 x 5 x 5 or 375 numbers from 1000 to 3999.

Including 4000, there will be 376 such numbers.

29 11823
Q:

Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

 A) 25200 B) 52000 C) 120 D) 24400

Explanation:

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)

= $\inline&space;{\color{Black}(7C_3{}\times&space;4C_2{})}$

= 210.

Number of groups, each having 3 consonants and 2 vowels = 210.

Each group contains 5 letters.

Number of ways of arranging 5 letters among themselves

= 5!

= 120

Required number of ways = (210 x 120) = 25200.

11 11382
Q:

A committee of 5 persons is to be formed from 6 men and 4 women. In how many ways can this be done when at least 2 women are included ?

 A) 196 B) 186 C) 190 D) 200

Explanation:

When at least 2 women are included.

The committee may consist of 3 women, 2 men : It can be done in$\inline&space;\fn_jvn&space;{\color{Black}&space;4C_{3}\times&space;6C_{2}}$  ways

or, 4 women, 1 man : It can be done in $\inline&space;\fn_jvn&space;{\color{Black}&space;4C_{4}\times&space;6C_{1}}$ ways

or, 2 women, 3 men : It can be done in $\inline&space;\fn_jvn&space;{\color{Black}&space;4C_{2}\times&space;6C_{3}}$ ways.

Total number of ways of forming the committees

= $\inline&space;\fn_jvn&space;{\color{Black}&space;4C_{2}\times&space;6C_{3}+4C_{3}\times&space;6C_{2}+4C_{4}\times&space;6C_{1}}$

= 6 x 20 + 4 x 15 + 1x 6

= 120 + 60 + 6 =186

10 8573
Q:

How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?

 A) 4050 B) 3600 C) 1200 D) 5040

Explanation:

'LOGARITHMS' contains 10 different letters.

Required number of words = Number of arrangements of 10 letters, taking 4 at a time.

= $\inline&space;{\color{Black}10P_4{}}$

= 5040.

17 7535
Q:

If the letters of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word ‘SACHIN’ appears at serial number :

 A) 601 B) 600 C) 603 D) 602

Explanation:

If the word started with the letter A then the remaining 5 positions can be filled in  5! Ways.

If it started with c then the remaining 5 positions can be filled in 5! Ways.Similarly if it started with H,I,N the remaining 5 positions can be filled in 5! Ways.

If it started with S then the remaining position can be filled with A,C,H,I,N in alphabetical order as on dictionary.

The required word SACHIN can be obtained after the 5X5!=600 Ways i.e. SACHIN is the 601th letter.