Elitmus
Exam
Numerical Ability
Permutation and Combination
There is a 3*3 grid,9squares are to be attached with stickers.There are3 red and 6 white stickers.each and every square needs a sticker. In how many ways the squares can be attached with stickers
Option
1)9c3
2)9c3+6c3
3)9c3/3!
4)(9c3+6c3)/3!
Read Solution (Total 14)
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- there are 9 boxes at start . now if we choose the red stickers then it would be 9C3 now we r left with 6 boxes and 6 stickers so it wil be 6c6.
ans is 9c3 * 6c6 =9c3 - 10 years agoHelpfull: Yes(43) No(0)
- 9C3*6C6=9C3
- 10 years agoHelpfull: Yes(4) No(2)
- 9C3*6C6=9C3
- 10 years agoHelpfull: Yes(2) No(3)
- 9C3
bcz if we sleact 3 out of 9 remaining 6 will be automaticaly taken by remaining 6
no need to slect remining 6 out of 6 - 10 years agoHelpfull: Yes(2) No(1)
- 9c3
for first 3 square 3 red ball can occupy 3 square out of 9 squares
thus it can be done in 9c3 ways
nxt 6 white ball can be done in 6c6 ways
ans will be 9c3*6c6=9c3
- 10 years agoHelpfull: Yes(2) No(1)
- Total ways will be = 9! / 3!*6!= 9C3
- 10 years agoHelpfull: Yes(2) No(1)
- there are 9 items
they can be arranged in 9 seats in 9! ways
but the red stickers are of same color and same for white stickers
therefore answer is 9!/6! 3! ie 9c3 - 10 years agoHelpfull: Yes(2) No(0)
- B)9C3+6C3
- 10 years agoHelpfull: Yes(1) No(8)
- 9c3*6c6=9c3
- 10 years agoHelpfull: Yes(1) No(1)
- 9c3 ways squares can be attached with stickers.
- 10 years agoHelpfull: Yes(1) No(2)
- select 3 out of 9 or 6 out of 3
9c3+6c3 - 10 years agoHelpfull: Yes(1) No(4)
- ans 1)9c3 ways
- 10 years agoHelpfull: Yes(1) No(1)
- there are 3X3 or 9 grids that are to be filled with 3 red and 6 white stickers. hence to atart filling we can start with either red stickers or white.. lets first fill with red stickers. so we choose 3 grids out of 9 by 9C3 and the rest are filled in one way by the rest 6 stickers.ans is 9C3*6C6=9C3. agn if we strt with whitethen simillarly ans is 9C6*3C3=9C6=9C3(nCr=nC(n-r))
- 10 years agoHelpfull: Yes(1) No(1)
- as there are 3 red and 6 white tickers which are f same type. therefore no of ways = 9!/6!*3! = 9C3/3! ways.
- 9 years agoHelpfull: Yes(0) No(0)
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