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Of the 120 people in a room, 3/5 are women. If 2/3 of the people are married, what is the maximum number of women in the room who could be unmarried? a)80 b)72 c)48 d)40 e)32

Q2. 3 friends are playing a game in which each person simultaneously displays one of three hand signs; a clenched fist, an open palm, or two extended fingers. How many different combinations of the signs are possible.?

Q2. 3 friends are playing a game in which each person simultaneously displays one of three hand signs; a clenched fist, an open palm, or two extended fingers. How many different combinations of the signs are possible.?

same signs for friends = 3(HHH, EEE, OOO) 2 same signs and one different sign = 2 * 3 = 6 (HHO, HHE, OOH, OOE, EEH, EEO) 3 different signs = 1 (HEO)

Of the 120 people in a room, 3/5 are women. If 2/3 of the people are married, what is the maximum number of women in the room who could be unmarried? a)80 b)72 c)48 d)40 e)32

Of the 120 people in a room, 3/5 are women. If 2/3 of the people are married, what is the maximum number of women in the room who could be unmarried? a)80 b)72 c)48 d)40 e)32

Q2. 3 friends are playing a game in which each person simultaneously displays one of three hand signs; a clenched fist, an open palm, or two extended fingers. How many different combinations of the signs are possible.?

Q2. 3 friends are playing a game in which each person simultaneously displays one of three hand signs; a clenched fist, an open palm, or two extended fingers. How many different combinations of the signs are possible.?

same signs for friends = 3(HHH, EEE, OOO) 2 same signs and one different sign = 2 * 3 = 6 (HHO, HHE, OOH, OOE, EEH, EEO) 3 different signs = 1 (HEO)

total 10

C

A list lost here.... Would be great if u could help me with this.

3 symbols.... and 3 people to choose.... Shouldn't the total symbols comb = 3x3x3 = 27...!
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|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

Q2. 3 friends are playing a game in which each person simultaneously displays one of three hand signs; a clenched fist, an open palm, or two extended fingers. How many different combinations of the signs are possible.?

same signs for friends = 3(HHH, EEE, OOO) 2 same signs and one different sign = 2 * 3 = 6 (HHO, HHE, OOH, OOE, EEH, EEO) 3 different signs = 1 (HEO)

total 10

C

A list lost here.... Would be great if u could help me with this.

3 symbols.... and 3 people to choose.... Shouldn't the total symbols comb = 3x3x3 = 27...!

Here we are talking about combinations and not permutations ... lets take the case of 3 friends and each choosing differnt symbol.

combinations = 1(no matter what each child picks, they will be picking the same 3 signs in different order) permutation = 3!

Here's the answer that I posted earlier. same signs for friends = 3(HHH, EEE, OOO) 2 same signs and one different sign = 2 * 3 = 6 (HHO, HHE, OOH, OOE, EEH, EEO) 3 different signs = 1 (HEO) total 10

Now take 3 different signs - HEO You can have 6 different types you can select(order matters) - HEO HOE EHO EOH OEH OHE Now if the order doesnt matter - then only 1 combination HEO as the rest of the 5 have the same alphabets but in different order.

Of the 120 people in a room, 3/5 are women. If 2/3 of the people are married, what is the maximum number of women in the room who could be unmarried? a)80 b)72 c)48 d)40 e)32

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