**How many bit strings of length 14 contains the following types (I) atmost three is? (ii) an equal number of 0s and 1s (iii) an odd number of 1s**

The **Answer to the Question**

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**Here's the Solution to this Question**

I. Atmost three

$C(n,r)=\frac{n!}{(n-r)!r!}\\ n=14\\ r≤3\\ =C(14,0)+C(14,1)+C(14,2)+C(14,3)\\ =1+14+91+364=470$

II. An equal number of 0s and 1s. This means that there will be 7 1s and 7 0s.

$n=14\\ r=7\\ =C(14,7)\\ =3432$

III. An odd number of 1s. This means that 1s can be 1,3,5,7,9,11,13.

$=C(14,1)+C(14,3)+C(14,5)+C(14,7)+C(14,9)+C(14,11)+C(14,13)\\ =14+364+2002+3432+2002+364+14\\ =8192$