Permutations and Combinations

A. 6
B. 9
C. 15
D. 18
Explanation:
This type of question becomes very easy when we assume three colour are red(R) blue(B) and Green(G).
We can choose any colour.
Now according to the statement 1 i.e.., codes can be generated by waving single flag of different colours, then number of ways are three i.e.., R.B.G from statement III three flags in different sequence of colours, then number of ways are six i.e.., RBG, BGR, GBR, RGB, BRG, GRB.
Hence total number of ways by changing flag = 3+ 6 +6 = 15

A. 63
B. 511
C. 1023
D. 15
Explanation:
Given that, the question paper consists of five problems. For each problem, one or two or three or none of the choices can be attempted.
Hence, the required number of ways = 45 – 1.
= 210 – 1 = 1024 – 1 = 1023

A. 400
B. 500
C. 600
D. 20
Explanation:
We can select one boy from 20 boys in 20 ways.
We select one girl from 25 girls in 25 ways
We select a boy and girl in 20 * 25 ways i.e., = 500 ways.

A. 25
B. 13
C. 40
D. 30
Explanation:
The word contains five consonants. Three vowels, three consonants can be selected from five consonants in ⁵C₃ ways, two vowels can be selected from three vowels in ³C₂ ways.
3 consonants and 2 vowels can be selected in ⁵C₂ . ³C₂ ways i.e., 10 * 3 = 30 ways.

A. ⁹C₂
B. ³C₂
C. ¹⁶C₂
D. ¹²C₂
Explanation:
Total number of balls = 9 + 3 + 4
Two balls can be drawn from 16 balls in ¹⁶C₂ ways.

A. 60
B. 360
C. 120
D. 240
Explanation:
The given digits are 1, 2, 3, 5, 7, 9
A number is even when its units digit is even. Of the given digits, two is the only even digit.
Units place is filled with only ‘2’ and the remaining three places can be filled in ⁵P₃ ways.
Number of even numbers = ⁵P₃ = 60.

A. 360
B. 60
C. 300
D. 180
Explanation:
The given digits are six.
The number of four digit numbers that can be formed using six digits is ⁶P₄ = 6 * 5 * 4 * 3 = 360.

A. (6!)2
B. 6! * ⁷P₆
C. 2(6!)
D. 6! * 7
Explanation:
We can initially arrange the six boys in 6! ways.
Having done this, now three are seven places and six girls to be arranged. This can be done in ⁷P₆ ways.
Hence required number of ways = 6! * ⁷P₆

A. 216
B. 243
C. 215
D. 729
Explanation:
Since each ring consists of six different letters, the total number of attempts possible with the three rings is = 6 * 6 * 6 = 216. Of these attempts, one of them is a successful attempt.
Maximum number of unsuccessful attempts = 216 – 1 = 215.

A. 165
B. 185
C. 205
D. 225
Explanation:
Total number of persons in the committee = 5 + 6 = 11
Number of ways of selecting group of eight persons = ¹¹C₈ = ¹¹C₃ = (11 * 10 * 9)/(3 * 2) = 165 ways.