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Number of possible cases = 10c2
=10!/(8!*2!)
=10*9/2
=45
Answer C
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How many possible cases are there such that 2 stocks are selected at random from 10 stocks?

A. 35
B. 40
C. 45
D. 50
E. 55


* A solution will be posted in two days.

Hi,
the Q is of COMBINATIONS but marked as PROBABILITY..
Please change it..

Method 1:
No. of ways of selecting two out of 10 stocks = 10C2 = 10!/(2!*8!) = 45

Method 2:
No. of ways of making pair of one stock with each of the remaining stock = 9
Total Such possible pairs (with Repetition) = 10*9 = 90
Since every case has been counted twice therefore total Unique pairs = 10*9 / 2 = 45

Method 3:
Pairs of first stock with each of the 9 stocks can be made in 9 ways

Pairs of Second stock with each of the remaining 8 stocks can be made in 8 ways (second has already made pair
with first so we will not make that pair again

Pairs of Third stock with each of the remaining 7 stocks can be made in 7 ways (Third has already made pair with first and second so we will not make those pairs again) .... and so on...

Total Such pairs = 9+8+7+6+5+4+3+2+1 = 45

Answer: option C
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How many possible cases are there such that 2 stocks are selected at random from 10 stocks?

A. 35
B. 40
C. 45
D. 50
E. 55

->10C2=10*9/2*1=45. Thus, the answer is C.
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MathRevolution
How many possible cases are there such that 2 stocks are selected at random from 10 stocks?

A. 35
B. 40
C. 45
D. 50
E. 55


* A solution will be posted in two days.

Hi,
the Q is of COMBINATIONS but marked as PROBABILITY..
Please change it..

Edited the tag. Thank you.




sir what is the difference between random and simultaneously. can we use 10c1x9c1 in random case? please explain
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