Explanation
Assuming, cards are drawn without replacement:
Total possible events =52C2
P(both red) =26C252C2
P(both king) =4C252C2
P(both red & king) =2C252C2
We use the basic addition rule,
P(both black or both queens) = P(both red) + P(both queens) - P(both black as well as queens)
Required probability =26C2+4C2−2C252C2
We have total digits 0,1,2,3,4.
We want to form odd number, then at last digit can become 1 or 3.
So, two number are fixed for last.
Now, we have option for 1st,2nd,3rd and 4th place are 3,3,2,1
So , number of getting odd numbers using given five distinct digit =3×3×2×1×2=36
Hence, this is the answer.
Given that,n+1C3=4.nC2
Then n=?
n+1C3=4.nC2
⇒(n+1)!3!(n+1−3)!=4.n!2!(n−2)!
⇒(n+1)n!3×2!(n−2)!=4.n!2!(n−2)!
⇒(n+1)3=4
⇒n+1=12
⇒n=12−1=11
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