Question

If a cricket team of 11 players is to be selected from 8 batsman, 6 bowlers, 4 all-rounder and 2 wicket-keepers, then the number of selections when at most 1 all-rounder and 1 wicket-keeper will play is
A.$^4{C_1}{.^{14}}{C_{10}}{ + ^2}{C_1}{.^{14}}{C_{20}}{ + ^4}{C_1}^2{C_1}{ + ^{14}}{C_{11}}$
B.$^4{C_1}{.^{15}}{C_{11}}{ + ^{15}}{C_{11}}$
C.$^4{C_1}{.^{15}}{C_{10}}{ + ^{15}}{C_{11}}$
D.$^4{C_1}{.^2}{C_1}^{14}{C_9}{ + ^{14}}{C_{11}}$

Answer 1

In the case of combination, we attach no importance to the order of things in a group. The number of permutations of n objects taken r at a time is determined by the following formula: P(n,r)=n! (n−r)! The code has 4 digits in a specific order, the digits are between 0-9.

Complete step-by-step answer:
Step1: Total number of players selected for cricket team =11
Number of batsman = 8
Number of bowlers = 6
Number of all-rounder = 4
number of wicket-keeper = 2
Arrangements when almost 1 all-rounder and 1 wicket-keeper.
Step2: when 1 all-rounder and 10 players from bowlers and batsman play number of ways
${}^4{C_1}{.^{14}}{C_{10}}$
When one wicketkeeper and 10 players from bowlers and batsman play
Number of ways = $^2{C_1}{.^{14}}{C_{10}}$
When one all-rounder and one wicket-keeper and 9 from the batsman and bowlers play
Number of ways = $^4{C_1}{.^2}{C_1}{.^{14}}{C_9}$
When all eleven players play from bowlers and batsman
Number of ways =$^{14}{C_{11}}$
∴ total number of selections = ${}^4{C_1}{.^{14}}{C_{10}}$+$^2{C_1}{.^{14}}{C_{10}}$+$^{14}{C_{11}}$= $^{14}{C_{11}}{ + ^4}{C_1}{.^2}{C_1}{.^{14}}{C_{11}}$
So (A) is the correct answer.

Note: A group or a selection, which can be formed by taking some or all of the number of objects irrespective of the order of their arrangement is called a combination.
$C(n,r)o{r^n}{C_r}$denoted as the number of combinations of n objects, taken r at a time.