wordings could be tricky sometimes ...but i still feel it cannot be 6P4
if the 4 choices are idli, dosa, utappa and wada ...problem wordings imply it does not matter who has ordered what ....coz it is not a different order ..
I am back with Probability...
Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?
A 1/5
B 2/5
C 3/5
D 4/5
E 3/8
I am back with Probability...
Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?
A 1/5
B 2/5
C 3/5
D 4/5
E 3/8
thats wrong !!! Try again..
one observation.. you guys seem to taken aback by mismatch in the actual answer the text in question stem.. plz.. read the lines below..
NEVER EXPECT GMAT (OR FOR THAT MATTER ANY OTHER APTITUDE TEST) TO BE EXPLICIT IN IT'S QUESTIONS... THE TEST IS NOT DESIGNED TO GAUGE CANDIDATE'S MATHEMATICAL SKILLS, BUT RATHER IT IS PRIMARILY DESIGNED TO GAUGE CANDIDATE'S ABILITY TO PICK THE RIGHT APPROACH IN THE LIMITED TIME AND REACH THE RIGHT ANSWER..
Dont wanna sound rude or anything, but I would really want to see a question in GMAT/CAT/GRE which explicitly asks the candidate to calculate a value using P&C; concepts.. hope that clarifies things.. ;)
have a good one!!
A family of 4 is at a restaurant with 6 choices
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Pure combinations ...internal pattern of choices does not matter
6C4 = 15 ..Option A
I sooo agree with you!! but the OA is 360.. using permutations...
and i have no clue why!! :-(
fyi - the ques is from a score800 test..
wordings could be tricky sometimes ...but i still feel it cannot be 6P4
.
.
..
In this question the key is words "nobody orders the same meal". This means order, although not mentioned, is to be kept in mind while solving this question.
.
.
.
how much time you took to solve this?
BTW... your approach is correct...but answer is wrong... only one thing you missed.
What are we finding here?
The probability of couple sitting together or not sitting together? ;)
..Let the couples be denoted as A1-A2 and B1-B2.
A1-A2 together = 4!2!
B1-B2 together = 4!2!
A1-A2 together and B1-B2 together = 3!2!2!
All A1-A2 together and/or all B1-B2 together = 2*4!2! - 3!2!2!
= 96-24 = 72
All possible arrangements without restriction = 5! = 120
Probability = 1 - (72/120)
= 2/5
answer = 1 - 2/5 = 3/5
I would have marked option B
I believe this question is too tough and time consuming to be asked in GMAT.. nevertheless, plz lemme know the OA ;)
nuttyvarun Sayscorrected the mistake committed above
One more for more practice:
A number sequence has 100 elements. Any of its elements (except for the first and last element) is equal to the product of its neighbors. The product of the first 50 elements, just as the product of all the elements is 27. What is the sum of the first and the second element?
a. 244/3
b. 10
c. 12
d. 28/3
e.1
Lengthy one:
There were ten friends four of whom are Sanjay, Salim, Reena and Teena. A team of six is to be formed such that Reena & Teena are never together but Sanjay and Salim are always together. How many teams can be made?
a.110
b.68
c.77
d.76
e.108
In this question the key is words "nobody orders the same meal". This means order, although not mentioned, is to be kept in mind while solving this question.
A-can order from 6 choices
B-has now only 5 choices to select from
C-has 4 after B has selected
D-has 3 choices after C has selected
Total - 6*5*4*3 = 360. So order came into play.
I hope this helps.
simple one...
A credit card number has 6 digits (between 1 to 9). The first two digits are 12 in that order, the third digit is bigger than 6, the forth is divisible by 3 and the fifth digit is 3 times the sixth. How many different credit card numbers exist?a)27.b)36.c)72.d)112.e)422.
ndhadha Sayswats the OA?
Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a second card is drawn. If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5 ?
A. 1/6
B. 1/5
C. 1/3
D. 2/5
E. 2/3
simple one...
A credit card number has 6 digits (between 1 to 9). The first two digits are 12 in that order, the third digit is bigger than 6, the forth is divisible by 3 and the fifth digit is 3 times the sixth. How many different credit card numbers exist?a)27.b)36.c)72.d)112.e)422.
Not sure if this approach is correct....
Possibilities for 3rd digit - 3
Possibilities for 4th digit - 3
Possibilities for 6th digit - 4
5th digit will always by 3 times 6th digit, so possibility - 1
Poss credit cards -> 3*3*4 = 36
IMO b) 36...
One more for practice:
In a blue jar there are red, white and green balls. The probability of drawing a red ball is 1/5. The probability of drawing a red ball, returning it, and then drawing a white ball is 1/10. What is the probability of drawing a white ball?
a)1/5.
b).
c)1/3.
d)3/10.
e).
Last one that i could solve today:
How many 5 digit numbers can be created if the following terms apply: the leftmost digit is even, the second is odd, the third is a non even prime and the fourth and fifth are two random digits not used before in the number?
a)2520
b)3150
c)3360
d)6000
e)7500
I am back with Probability...
Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?
A 1/5
B 2/5
C 3/5
D 4/5
E 3/8
One more for more practice:
A number sequence has 100 elements. Any of its elements (except for the first and last element) is equal to the product of its neighbors. The product of the first 50 elements, just as the product of all the elements is 27. What is the sum of the first and the second element?
a. 244/3
b. 10
c. 12
d. 28/3
e.1
