Waste less time on Facebook — follow Brilliant.
×

please help..permutations..

a person wishes to make up as many different parties out of his 20 friends such that each party consists of same no of persons.the number of persons he should invite at a time is equal to....plzzzz answer this fast

Note by Varnika Chaturvedi
3 years, 11 months ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

Think about it like this. If he invites only one person to a party, he can only give 20 parties. Similarly, if he invites everyone except for one person to a party, he can also only give 20 parties. So what if he decides to invite two people to each party? Then how many parties can he give? Is this more than with one person? Tim Vermeulen · 3 years, 11 months ago

Log in to reply

we can apply the max value of nCr,here as n is even max value is 20C10,re answer 184756 Nayan Pathak · 3 years, 10 months ago

Log in to reply

This qn is actually a combination qn, not a permutation since the order of the friends is not important.

If he invites n number of ppl to the party, the possible number of parties is \(^{20} C_n = \frac{20\times19\times....\times(20-n)}{n!}\)

Lets see the pattern

1 ppl parties: \(^{20}C_1 = 20\)

2 ppl parties: \(^{20}C_2 = 190\)

3 ppl parties: \(^{20}C_3 = 1,140\)

4 ppl parties: \(^{20}C_4 = 4,845\)

5 ppl parties: \(^{20}C_5 = 15,504\)

6 ppl parties: \(^{20}C_6 = 38,760\)

7 ppl parties: \(^{20}C_7 = 77,520\)

8 ppl parties: \(^{20}C_8 = 125,970\)

9 ppl parties: \(^{20}C_9 = 167,960\)

10 ppl parties: \(^{20}C_{10} = 184,756\)

11 ppl parties: \(^{20}C_{11} = 167,960\)

12 ppl parties: \(^{20}C_{12} = 125,970\)

13 ppl parties: \(^{20}C_{13} = 77,520\)

14 ppl parties: \(^{20}C_{14} = 38,760\)

15 ppl parties: \(^{20}C_{15} = 15,504\)

16 ppl parties: \(^{20}C_{16} = 4,845\)

17 ppl parties: \(^{20}C_{17} = 1,140\)

18 ppl parties: \(^{20}C_{18} = 190\)

19 ppl parties: \(^{20}C_{19} = 20\)

20 ppl parties: \(^{20}C_{20} = 1\)

As we see, we have the most number of parties if we invite 10 friends at a time. We can have 184, 756 different parties!!! Saad Haider · 3 years, 11 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...