Question

Three letters are chosen at random from the word HEART and arranged in a row. Find the
probability that:
both vowels are chosen

Answer 1

The no. of ways in which three letters can be chosen from 5 letters of HEART is 5C3 = 10

If both vowels are to be chosen, there remains a single place for other 3 consonants letters, So, the no. of ways to make a combination that includes both vowels = 3C1 = 3

Hence, the probability of choosing both the vowels = 3/10

Answer 2

got it!
another one.. Three men and three women are to be randomly seated in a row. Find the probability that
both the end places will be filled by women.

Answer 3

Okay. The no. of ways in which 3 men and 3 women can sit in a row is , 6! / (3! *3!) = 20

When 2 end places are already occupied by women, then there are remaining 4 places to be filled by 3 men and 1 Women. So, the no. of ways in this case= 4! / (3!) = 4

Hence, the probability = 4/20 = 1/5

Answer 4

its okay i got it!
um number of way 3 men and 3 women sit in a row is 6!/0! ?

Answer 5

Nope..

The no. of ways in which 'N' nos. of things can be arranged, in which one item repeat P times , another repeat Q times is,

N! / (P! * Q!)...

[ N! is for unrepeated case where every object is different]

Answer 6

three letters are chosen from 5 letter word.. So it would be 5C3..

Probability both vowels r chosen.. 2C2
and for choosing the third letter, 3C1..

so the Probability of the event = (2C2 * 3C1) / 5C3..

Answer 7

hm why isnt it permutation but combination?

Answer 8

because the orders in which you choose your letters is irrelevant.

HEA and AEH are essentially the same, so the orders don't matter at all

And, by the way.. don't I deserve a medal here.. So many people I've helped don't give me medal at all.. am I that aloof?