Permutations & Combinations - Questions & Discussions

MBA CAT 2022
832
We play the coin tossing game in which if tosses match, I get both coins; if they differ, you get both. If you have m coins and I have n, what is the expected length of the game (i.e., the number of tosses until one of us is wiped out)?
833
@didroo you have to select a pair of shoe so that can be done only in 10 ways as there are 10 pairs available and once a pair is selected you can select one shoe from remaining 18 shoes i.e. 18 ways to do it. that's why 10*18
834
@srikanthp09 ans 540 is possible only if we assume that 3 boys are identical.. though it sounds meaningless but thats how the solution is coming.
835

In a pack of cards of 52,if a card is randomly picked and its number is noted down and place the card in the pack.This process is repeated 9 times .what is the remainder when the sum is divided by 17

836
@somu6211 said:
In a pack of cards of 52,if a card is randomly picked and its number is noted down and place the card in the pack.This process is repeated 9 times .what is the remainder when the sum is divided by 17
6?
837
@maddy2807 said:
6?
kaise? can u enlighten!
838
@naarto said:
kaise? can u enlighten!
bhai Not at all sure for the ans... I hv taken general cases and added up the Numbers.... took ace as 1 and king as 13...
correct me if i m wrng?
839

the question shld have been the remainder when no of ways of drawing the cards in such way divided by 17. 😃 Bas aisa kahi dekha lag raha mujhe isliye 😁

840
@naarto question wahi hai jo aap bol rahe ho..yeh question kaha aya hai pata hai mujhe..bt shayad bola nahi jasakta hai abhi uska source.. :P
841
@Squib said:
@naarto question wahi hai jo aap bol rahe ho..yeh question kaha aya hai pata hai mujhe..bt shayad bola nahi jasakta hai abhi uska source..
aisa bhi mat bolna :P
842
A few number of four-letter words are formed by using 17 consonants and 5 vowels. How many words out of them will have 2 different vowels in the middle and a consonant at each end?

843
@ajeetaryans said:
A few number of four-letter words are formed by using 17 consonants and 5 vowels. How many words out of them will have 2 different vowels in the middle and a consonant at each end?
5440?
844
@ajeetaryans said:
A few number of four-letter words are formed by using 17 consonants and 5 vowels. How many words out of them will have 2 different vowels in the middle and a consonant at each end?


vowel can be arranged in 5c2 * 2 ! = 20 ways

consonant can be same or different .. so 17^2 = 289

so in total 5780 ways

845
Two students play a game based on the total roll of two standard dice. Student A says that a 12 will be rolled first. Student B says that two consecutive 7s will be rolled first. The students keep rolling until one of them wins. What is the probability that A will win?

846
@ajeetaryans said:
Two students play a game based on the total roll of two standard dice. Student A says that a 12 will be rolled first. Student B says that two consecutive 7s will be rolled first. The students keep rolling until one of them wins. What is the probability that A will win?
36/71?
847
@maddy2807

Ans shud be 1 as 52*9mod 17 gives 1 as remainder.
848
@@ankur_tiger options for that question

1)0

2)1

3)3

4)9

I think 1 is the right answer

849

@somu6211 dost so u mean ans is option b right ?
850
This is a very easy question & i already know the solution but i am confused about another approach(May be my concepts are not clear)..ďťż

Q. There are 5 boys & 6 girls. A committee of 4 is to be selected so that it must consist of at least 1 boy & 1 girl??

Conventional way to solve this is by making 3 cases:
1 boy & 3 girls
2 boys & 2 girls
3 boys & 1 girl
& add them up

5c1X6c3 + 5c2X6c2 + 5c3X6c1 = 310

What's wrong if we consider one boy & one girl first receptively and then select the other 2 members from the remaining 9 candidates, which would lead to this.

5c1 X 6c1 X 9c2 = 1080

I know this is dumb but i'm stuck!!!

851
@farziPandit said:
This is a very easy question & i already know the solution but i am confused about another approach(May be my concepts are not clear)..Q. There are 5 boys & 6 girls. A committee of 4 is to be selected so that it must consist of at least 1 boy & 1 girl??Conventional way to solve this is by making 3 cases:1 boy & 3 girls2 boys & 2 girls 3 boys & 1 girl& add them up5c1X6c3 + 5c2X6c2 + 5c3X6c1 = 310What's wrong if we consider one boy & one girl first receptively and then select the other 2 members from the remaining 9 candidates, which would lead to this.5c1 X 6c1 X 9c2 = 1080I know this is dumb but i'm stuck!!!
@pendyal @keviv @GauravShah @maddy2807 @naarto @naga25french @Aizen @Estallar12 @chillfactor
Can Anyone of you help me out with this one??