If $\binom{189}{35} + \binom{189}{x} = \binom{190}{x}$,then $x = \dots$

The number of ways in which a cricket team of $11$ members can be formed out of $6$ batsmen,$6$ bowlers,$4$ all-rounders and $4$ wicket keepers by selecting at least $4$ batsmen,at least $3$ bowlers,at least $2$ all-rounders and only $1$ wicket keeper is:

If $4$ red balls and $5$ green balls are selected from $n$ balls,and the sum of both selections is greater than ${}^{n+1}C_4$,then the value of $n$ is equal to:

There are two urns. Urn $A$ has $3$ distinct red balls and urn $B$ has $9$ distinct blue balls. From each urn,two balls are taken out at random and then transferred to the other. The number of ways in which this can be done is

In an examination,a question paper consists of $12$ questions divided into two parts,i.e.,Part $I$ and Part $II$,containing $5$ and $7$ questions,respectively. $A$ student is required to attempt $8$ questions in all,selecting at least $3$ from each part. In how many ways can a student select the questions?

From all the English alphabets,five letters are chosen and are arranged in alphabetical order. The total number of ways,in which the middle letter is $M$,is: