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The correct option is (a) 9

Easiest EXPLANATION: A hexagon has 6 sides. We obtain the diagonals by JOINING the vertices in pairs.

Total number of sides and diagonals = ^6C2 = \(\frac{6 * 5}{2 * 1}\) = 5×3 = 15. This INCLUDES its 6 sides ALSO. So, Diagonals = 15 – 6 = 9. Hence, the number of diagonals is 9.

Correct option is (d) 37

The explanation: Let’s CONSIDER the given STRING is CLEAN, so set of string of length 1 = {C,L,E,A,N} ; cardinality of set = 5 set of string of length 2 = {CL,EE,EA,NN}, set of string of length 3 = {CLE,LEE,EAN}, set of strings of length 4 = {CLEN,LEAN}, set of strings of length 5 = {CLEAN} and set of string of length 0 = {} and we cannot have any substring of length 6 as given string has only 5 length. So total no of SUBSTRINGS are possible = 0 length substring + 1 length substring + 2 length substrings +3 length substrings + 4 length substrings + 5 length substrings = 1 + 5 + 4 + 3 + 2 + 1 = 16 means for 1 length string to n length substrings it will sum of the n natural no from 1 to n.

so 1+2+3+…+n = \(\FRAC{n(n+1)}{2}\) so total no substrings possible = 0 length strings + \(\frac{n(n+1)}{2}\) = 1+ \([\frac{n(n+1)}{2}]\) so total no of substrings possible in n length string (All length inclusive )= 1 + \([\frac{n(n+1)}{2}] = \frac{8(8+1)}{2}\) = 37.

Right answer is (b) 18

The BEST I can explain: Consider three buckets red, WHITE and blue and we WANT the total number of balls such that each bucket contain at least one ball. Now consider the state of picking up a ball without replacement : (normally you consider the worst CASE scenario in these cases) Starting 10 balls all are red and thus goes to bucket NAME Red. Now again picking up the ball gives 7 balls which are of same colour and put all of them in a bucket named White. The next pick will definitely be of different colour thus: we picked 10 + 7 + 1 = 18.

The correct option is (d) 1260

For explanation: Dolls are different but the boxes are identical. If none of the boxes is to remain empty, then we can pack the dolls in one of the FOLLOWING ways:

Case i. 2, 2, 2, 1, 1

Case II. 3, 3, 1, 1

Case i: Number of ways of achieving the first option 2, 2, 2, 1, 1. Two dolls out of the 8 can be selected in ^8C2 ways, another 2 out of the remaining 6 can be selected in ^6C2 ways, another 2 out of the remaining 4 can be selected in ^4C2 ways and the last two dolls can be selected in ^1C1 ways each. However, as the boxes are identical, the two different ways of selecting which box holds the first two dolls and which one holds the second set of two dolls will look the same. Hence, we need to divide the result by 2. Therefore, TOTAL number of ways of achieving the 2, 2, 2, 1, 1 is = (^8C2 * ^6C2 * ^4C2* ^1C1* ^1C1) / 2 = 1260.

Correct answer is (d) 24

To elaborate: Let a1 be the number of games played until day 1, and so on, AI be the no games played until i. Consider a sequence like a1,a2,…a30where 1≤ai≤45, ∀ai. Add 14 to each element of the sequence we get a new sequence a1+14, a2+14, … A30+14where, 15 ≤ ai+14 ≤ 59, ∀ai. Now we have two sequences 1. a1, a2, …, a30 and 2. a1+14, a2+14, …, a30+14. having 60 elements in total with each elements taking a value ≤ 59. So according to pigeon hole principle, there must be at LEAST two elements taking the same value ≤59 i.e., ai = aj + 14 for some i and j. THEREFORE, there exists at least a period such as aj to ai, in which 14 matches are played.

The correct CHOICE is (c) 11

The explanation is: The question REQUIRES you to find number of the outcomes in which at most 2 coins TURN up as heads i.e., 0 coins turn heads or 1 coin turns head or 2 coins turn heads. The number of outcomes in which 0 coins turn heads is ^4C0 = 1 outcome. The number of outcomes in which 1 coin turns head is ^4C1 = 6 outcomes. The number of outcomes in which 2 coins turn heads is,

^4C2 = 15 outcomes. Therefore, total number of outcomes = 1 + 4 + 6 = 11 outcomes.

Right option is (b) 5

To elaborate: With 2 elements pairs which give sum as 7 = {(1,6), (2,5), (3,4), (4,3)}. So choosing 1 ELEMENT from each group = 4 elements (in worst case 4 elements will be either {1,2,3,4} or {6,5,4,3}). Now using pigeonhole principle = we NEED to choose 1 more element so that sum will definitely be 7. So NUMBER of elements must be 4 + 1 = 5.

Correct choice is (b) 2

The best explanation: Suppose each of the 267 members of the group has at least 1 friend. In this case, each of the 267 members of the group will have 1 to 267-1=266 friends. Now, consider the numbers from 1 to n-1 as holes and the n members as pigeons. Since there is n-1 holes and n pigeons there MUST exist a hole which must CONTAIN more than one pigeon. That means there must exist a number from 1 to n-1 which would contain more than 1 member. So, in a group of n members there must exist at least TWO persons having equal number of friends. A similar case occurs when there exist a PERSON having no friends.

Right option is (B) 27384

Explanation: There are 16 × 15 × 14 + 14 × 13 × 12 × 11 = 27384 WAYS to choose from a student council.

The correct option is (d) 14

For EXPLANATION I WOULD say: Given 12 RED and 12 blue socks so, in order to take out at least 2 blue socks, first we need to take out 12 shocks (which might end up red in worst case) and then take out 2 socks (which would be definitely blue). THUS we need to take out total 14 socks.

The correct answer is (a) 10000000

Best explanation: By the BASIC COUNTING principle, the TOTAL NUMBER of PIN numbers Amit can choose is 10 × 10 × 10 × 10 × 10 × 10 × 10 = 10,000000.

The correct answer is (d) 456976000

To ELABORATE: 10^3 × 2^64 = 456976000 possible codes are FORMED for the SAFE with the alphanumeric digits.

Right option is (c) 829440

Best explanation: There are FOUR GROUPS of BOOKS which can be arranged in 4! different ways. Among those books, two are Geography books, FIVE are Natural Sciences books, three are History books and four are Mathematics books. Therefore, there are 4! × 2! × 5! × 3! × 4! = 829440 ways to arrange the books.

Correct choice is (d) 10500

Best EXPLANATION: By the BASIC Counting Principle, the number of different choices is 10 × 15 × 7 = 10500.

Correct OPTION is (a) 16807

Easy EXPLANATION: 7^5 = 16807 ways of making the numbers consisting of five DIGITS if REPETITION is allowed.

Right option is (d) 59406880

Explanation: The first letter must be a vowel, so there are 5 choices. The second letter can be any one of 26, the THIRD letter can be any one of 26, the fourth letter can be any one of 26 and fifth and sixth letters can be any of 26 choices. The last letter must be an A, so there is only 1 CHOICE. By the basic COUNTING principle, the number of ‘words’ is 5 × 26 × 26 × 26 × 26 × 26 × 1 = 59406880.

Right option is (C) 4500

To elaborate: The thousands digit cannot be zero, so there are 9 choices. There are 10 possibilities for the hundreds digit and 10 possibilities for the tens digit. The units digit can be 0, 2, 4, 6 or 8, so there are 5 choices. By the basic counting principle, the NUMBER of EVEN five digit WHOLE numbers is 9 × 10 × 10 × 5 = 45,00.