In Cricket League, in first round every team plays a match with every other team. 9 teams participated in the Cricket league. How many matches were played in the first round?
Answer: A
Every pair of teams meets once, and the order within a pair is irrelevant.
\[ {}^{9}C_{2} = \frac{9 \times 8}{2} = 36 \]
Counting ordered pairs would give 72, treating "A versus B" and "B versus A" as different matches.
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From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?
Answer: D
"At least 3 men" among five people splits into three cases.
\[ 3M, 2W: \ {}^{7}C_{3} \times {}^{6}C_{2} = 35 \times 15 = 525 \]
\[ 4M, 1W: \ {}^{7}C_{4} \times {}^{6}C_{1} = 35 \times 6 = 210 \]
\[ 5M, 0W: \ {}^{7}C_{5} = 21 \]
\[ 525 + 210 + 21 = 756 \]
The cases are mutually exclusive, so the counts add. Here the complement would need just as much work, since "fewer than 3 men" also has three cases.
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In how many ways can the letters of the word INDIA be arranged, such that all vowels are never together?
Answer: B
INDIA has five letters with I twice, and its vowels are I, I, A.
\[ \text{total} = \frac{5!}{2!} = 60 \]
With the three vowels as one block there are three units (block, N, D):
\[ 3! \times \frac{3!}{2!} = 6 \times 3 = 18 \text{ with the vowels together} \]
\[ 60 - 18 = 42 \]
The repeated I forces a division by \( 2! \) in both counts.
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There are 20 points in a plane, how many triangles can be formed by these points if 5 are colinear?
Answer: A
A triangle needs three points, but three collinear points give no triangle.
\[ \text{all triples} = {}^{20}C_{3} = 1140 \]
\[ \text{collinear triples} = {}^{5}C_{3} = 10 \]
\[ 1140 - 10 = 1130 \]
Only the five collinear points can produce a degenerate triple, so just one subtraction is needed.
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A person ordered 5 pairs of black socks and some pairs of brown socks. The price of a black pair was thrice that of a brown pair while preparing the bill the bill clerk inter-changed the number of black and brown pairs by mistake which increased the bill by 100% what was the number of pair of brown socks in the original order?
Answer: D
Let the bought x pairs of brown socks and the price of each brown pair be Y.
Then total cost = 5x3Y+xy
Changed cost = 5xY+x*3Y
According to the question .
5y+3xy -(15y+xy)/15y+xy x 100% = 100%
=> 5y+3xy-15y-xy/ 15y+xy x100% = 100%
=> 5y+3xy = 15y+xy+15y+xy
=> 5y+3xy = 30y+2xy
=> xy= 25y
=> x=25.
Hence the original pair of brown socks = 25.
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In a society, it is customary for friends of the same sex to hug and for friends of opposite sex to shake hands when they meet. A group of friends met in a party and there were 24 hand shakers. Which one among the following numbers indicates the possible number of hugs?
Answer: C
Let there be \( m \) men and \( f \) women. Handshakes happen only between opposite sexes and hugs only within a sex.
\[ mf = 24, \qquad \text{hugs} = {}^{m}C_{2} + {}^{f}C_{2} \]
Testing the factor pairs of 24:
So 21 is attainable, with four of one sex and six of the other.
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In how many ways can the letters of the word LEADER be arranged ?
Answer: C
Correct answer: 360
The word has 6 letters, of which E occurs twice and the rest once each.
\[\frac{6!}{2!} = \frac{720}{2} = 360\]
Dividing by 2! removes the arrangements that merely swap the two identical E’s.
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A bank has 6 digit account number with no repetition of digits within a account number. The first and last digit of the account numbers is fixed to be 4 and 7. How many such account numbers are possible?
Answer: D
The first and last digits are fixed at 4 and 7, so only the four middle positions are free - and they must avoid 4 and 7, leaving eight digits.
\[ 8 \times 7 \times 6 \times 5 = 1680 \]
Repetition is forbidden, so each position has one fewer choice than the last.
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Twelve people from a club, by picking lots. One of them will host a dinner for all once in a month. The number of dinners a particular member has to host in one year is
Answer: D
Each month's host is drawn by lot from all twelve members, so the draws are independent and a member's chance does not depend on whether they have hosted before.
Over twelve months a particular member might be drawn every time, never, or anything in between - the count is not fixed.
So the answer cannot be determined. It would be exactly one only if the lots were drawn without replacement.
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In how many ways can the letters of the word 'LEADER' be arranged?
Answer: C
LEADER has six letters with E appearing twice.
\[ \frac{6!}{2!} = \frac{720}{2} = 360 \]
The division by \( 2! \) removes the double-counting from swapping the two identical E's, which changes nothing visible.
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