Question

18) Read the Case study given below and attempt any 4 sub parts:
A state cricket authority has to choose a team of 11 members, to do it so the
authority asks 2 coaches of a government academy to select the team members
that have experience as well as the best performers in last 15 matches. They can
make up a team of 11 cricketers amongst 15 possible candidates. In how many
ways can be the final eleven be selected from 15 cricket players if:

i) there is no restriction
a. 1365
b. 2365
c. 1465
d. 1375
ii) one of them must be included
a. 1002
b. 1003
c. 1001
d. 1004
iii)one of them, who is in bad form, must always be excluded
a. 480
b. 364
c. 1365
d. 640
iv) Two of them being leg spinners, one and only one leg spinner must be
included?
a. 2C1 ×
13C10
b. 2C1 ×
10C13
c. 1C2 ×
13C10
d. 2C10 ×
13C10

v) If there are 6 bowlers, 3 wicket-keepers, and 6 batsmen in all. The
number of ways in which a team of 4 bowlers, 2 wicket–keepers and 5
batsmen can be chosen.

Answer 1

Answer :-

i) If there are no restrictions :-

Here, we have to select 11 players out ot 15 and there are no restrictions and here the order of the players doesn't matter. So we will here apply combination :-

⇒ $\sf ^nC_r = \dfrac{n!}{(n-r)! r!}$

⇒ $\sf ^{15} C_{11} = \dfrac{15!}{(15-11)! 11!}$

⇒ $\sf ^{15} C_{11} = \dfrac{15!}{11! 4!}$

⇒ $\sf ^{15} C_{11} = \dfrac{15 \times 14 \times 13 \times 12 \times \cancel{11!}}{\cancel{11!} 4!}$

⇒ $\sf ^{15} C_{11} = \dfrac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1}$

⇒ $\sf ^{15} C_{11} = 1365$

If there is no restriction, then possible ways to select 11 players from 15 are 1365.

ㅤ

ㅤ

ii) One of them must be included :-

If one player must always be included, then we have to select 10 players from 14 -

⇒ $\sf ^{14} C_{10} = \dfrac{14!}{(14-10)!10!}$

⇒ $\sf ^{14} C_{10} = \dfrac{14!}{4!10!}$

⇒ $\sf\dfrac{14 \times 13 \times 12 \times 11 \times \cancel{10!}}{\cancel{10!} 4!}$

⇒ $\sf ^{14} C_{10} = \dfrac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2}$

⇒ $\sf ^{14} C_{10} = 1001$

Possible ways to select the players if one them must be included are 1001.

ㅤ

ㅤ

iii) One of them, who is in bad form, must always be excluded.

If one of them must always be excluded, then we have to select 11 players out of 14 -

⇒ $\sf ^{14} C_{11} = \dfrac{14!}{(14-11)! 11!}$

⇒ $\sf ^{14} C_{11} = \dfrac{14!}{11! 3!}$

⇒ $\sf ^{14} C_{11} = \dfrac{14 \times 13 \times 12 \times \cancel{11!}}{\cancel{11!} 3!}$

⇒ $\sf ^{14} C_{11} = \dfrac{ 14 \times 13 \times 12}{ 3 \times 2 \times 1}$

⇒ $\sf ^{14} C_{11} = 364$

Possible ways to select 11 players out of 15 if one must always be excluded are 364.

ㅤ

ㅤ

iv) Two of them being leg spinners, one and only one leg spinner must be included -

Let's first find out possible ways to select players which are not leg spinner -

There are two leg spinners out of 15 and one players must be leg spinner.

So, we have to select 10 players out of 13 -

⇒ $\sf ^{13}C_{10}$

Now, there are 2 leg spinners and we have to select 1 so possible ways :-

⇒ $\sf ^2C_1$

Total possible ways to select 11 players out of 15 out of which one must be leg spinner out of 2 are ¹³C₁₀ x ²C₁

ㅤ

ㅤ

v) If there are 6 bowlers, 3 wicket-keepers, and 6 batsmen in all. The

number of ways in which a team of 4 bowlers, 2 wicket–keepers and 5

batsmen can be chosen.

Possible ways to choose 4 bowlers out of 6 -

⇒ $\sf ^6C_4 = 3$

Possible ways to select 5 batsmen out of 6 -

⇒ $\sf ^6C_5 = 6$

Possible ways to select 2 wicket keepers out of 3 -

⇒ $\sf ^3C_2 = 3$

Total ways = 6 × 3 × 3 = 54

Total ways to select 4 bowlers, 2 wicket–keepers and 5 batsmen out of 6 bowlers, 3 wicket-keepers, and 6 batsmen in all are 54.

Answer 2

Answer:

$\sf{\green{\underline{\underline{\orange{Answer :-⬆}}}}}$

IF ELSE IF LADDER:-The if else if ladder is an extension of the if else conditional branching statement. It is used when we have to check multiple conditions before taking a decision. The if else ladder provides a proper skeleton to check multiple conditions, where every condition has a block of statement within itself.

if(condition expression)

{

Java Statement 1;

-----------------------

}

else if(condition expression)