A synchronized diving competition will feature 25 synchr

We need atleast 25 different combinations of divers from say 'N' divers.

So, NC2>25

N*(N-1)/2 > 25
N*(N-1) > 50

N = 8 satisfies and is the minimum value.

Option C

Since each of the 25 jumps features should feature a different combination of divers, Lets start from the answer options,

(A) 6 -- 6C2 = 15 < 25
(B) 7 -- 7C2 = 21<25
(C) 8 - 8C2= 28>25 . Hence our answer option.


ANSWER OPTION C

Rephrasing the question, we need to find out a number, such that choosing 2 divers from the pool, will provide more than 25 combinations.
Thus, nC2 > 25.

If, we apply options, 6C2 = 15; 7C2 = 21; 8C2 = 28.
Thus C is the correct answer.

For minimum 25 jumps of 2 different combinations of swimmers, the minimum is 6.
No of swimmers taking 1st jump: 6
No of swimmers taking 2nd jump: 5
No of possible pairs = 6*5 = 30

8!/(2!*6!)=28 different jumps which is greater than 25. So the answer is "C"

8!/(2!*6!)=28 which is greater than 25.
so the answer is C.

A synchronized diving competition will feature 25 synchronized jumps of 2 divers each. At a minimum, how many divers are needed to ensure that each of the 25 jumps features a different combination of divers?

Number of divers needed = n
Combinations from n divers = nC2

We need a combination which gives 25 or more divers.

For n= 6, nC2 = 6C2 = 15
For n= 7, nC2 = 7C2 = 21
For n= 8, nC2 = 8C2 = 28

Therefore, a minimum of 8 divers are needed to ensure that each of the 25 jumps features a different combination of divers

Choice C is the answer.

A synchronized diving competition will feature 25 synchronized jumps of 2 divers each. At a minimum, how many divers are needed to ensure that each of the 25 jumps features a different combination of divers?

(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

Solution :

In this question 25 Synchronized jumps are of 2 divers each , so lets go in steps option by option :

1) Option A :Lets say we have 6 divers , then possible pairs of 2 divers can be 6c2 = 6*5/2 = 15 which is less than number of jumps 25


2) Option B : Lets say we have 7 divers , then possible pairs of 2 divers can be 7c2= 7*6/2 = 21 which is less than number of jumps 25

3) Option C : lets say we have 8 divers , then possible pairs of 2 divers can be 8c2 = 8*7/2= 28 Which is greater than
number of jumps 25 and thus will ensure that each of the 25 jumps features a different combination of divers


Thus answer is c ) 8

25 synchronised jumps mean that there would be 25 patterns made by 2 divers.
This would also represent an unordered and a without replacement condition. Which is given by the formula = nC2

We have to find the minimum number of n(number of divers) which would correspond to 25 sync patterns.

Trying with the answer choices -
1. 6C2 gives 15 sync patterns
2. 7C2 gives 21 sync patterns which is still less than 25
2. 8C2 gives 28 sync patterns

Therefore, minimum number of 8 divers would be required to form 25 sync patterns.

Answer - option 'C'

Let number of divers be x
Hence no. Of distinct combinations of 2 = xC2
Hence we need to find minimum value of x such that xC2>25
=>x=8, as 8C2=28

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All we need to do here is find out n for which nC2 will be >= 25.

(A) 6C2 = 15 (less than 25 -> OUT)
(B) 7C2 = 21 (less than 25 -> OUT)
(C) 8C2 = 28 Correct!

Thus, the minimum value of n should be 8! (C)
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Let us take 8 Divers
|||| ||||
the first diver will have 7 ways of combining to form a pair with the rest
ie if you take D1 D2 D3 D4 D5 D6 D7 D8
then D1 can combine with D2 , D1 with D3 and so on till D1 with D8 so total 7 ways
then D2 can combine with D3, D2 with D4 , So total 6 ways
D3 with D4, and so on , SO total 5 ways
D4 with D5, D5 with D6, so on total 4 ways
D5 with D6 and so on till 3 ways
D6 with D7, D6 with D8 => 2 ways
D7 with D8 => 1 ways
total = 7 + 6+ 5+ 4+ 3+ 2 + 1 => > 25

now let us take total divers = 7..no need to do calculation again
ot will be (n-1) ways
6 + 5 + 4+ 3+2+1 = 20 ways less than 25
and hence minimum required is 8 divers and hence answer is C

so a minimum of nC2 = 25
or\(\frac{ n(n-1)}{2}\)>= 25
n(n-1)>=50
8 is the minimum value
8*7=56>50
C IMO
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Answer is c
8
(1,2),(1,3),(1,4)......=7
(2,3),(2,4),.....=6
(3,4),....=5
same way
total jumps will be= 7+6+5+4+3+2+1=28 jumps minimum number will be 8 drivers

This can be solved using Permutations and Combination.
2 divers are needed for each jump and 25 such jumps need to be performed.

The number of ways of selecting two items from a set of n items is nC2.

Plug-in options { Start from middle and move above or below depending on approach towards answer}
C. 8 => 8C2 = 28 {More}
B. 7 => 7C2 = 21 {Less}

Hence the least number of divers required is 8
IMO C is the Answer

OA is C i,e., minimum of 8 divers are needed

Let no. of divers = n
we need to select 25 distinct pairs from 'n' people = \(C^2_n\) = \(\frac{n(n-1)}{2}\)

now, \(\frac{n(n-1)}{2}\) = 25

= \(n^2-n-50 = 0\)

n = \(\frac{1 + \sqrt{201}}{2}\)

n = \(\frac{1 + 14.xx}{2}\)
n=7.xx

so min. 'n' needed = 8

IMO C