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The manager of a football team of eleven players wants to take photographs of his team, six players at a time. He can choose anyone out of four renowned photographers to take the photographs. In how many ways can the photographs be taken, if the Captain and the Vice-Captain of the team are always included in the photographs? Also find out in how many ways can the photographs be taken, if the Captain and the Vice-Captain of the team are never included in the photographs?

The manager of a football team of eleven players wants to take photographs of his team, six players at a time. He can choose anyone out of four renowned photographers to take the photographs. In how many ways can the photographs be taken, if the Captain and the Vice-Captain of the team are always included in the photographs? Also find out in how many ways can the photographs be taken, if the Captain and the Vice-Captain of the team are never included in the photographs? Correct Answer <p>(<sup>9</sup>C<sub>4</sub> × 6! × 4); (<sup>9</sup>P<sub>6</sub> × 4)</p>

Calculation:

⇒ If the captain and the vice-captain are always included, the number of ways is 9C4 × 6!.

But, these photographs can be taken by the photographers in 4 different ways. 

⇒ So, the required number of ways is 9C4 × 6! × 4.
If the captain and the vice-captain are never included, the number of ways is 9C6 × 6!.

But, these photographs can be taken by the photographers in 4 different ways.

⇒ So, the required number of ways is 9C6 × 6! × 4 = 9P6 × 4.

Additional Information

⇒The total number of permutations on a set of n distinct objects is given by n! and is denoted as nPn = n!

⇒ The total number of permutations on a set of n objects taken r at a time is given by nPr = n!/(n - r)!

⇒ The number of ways of arranging n objects of which r is the same is given by n!/ r!

⇒ If we wish to arrange a total of n objects, out of which ‘p’ are of one type, q of second type are alike, and r of a third kind is the same, then such a computation is done by n!/p!q!r!

⇒ Almost all permutation questions involve putting things in order from a line where the order matters. For example, ABC is a different permutation to ACB.

⇒ The number of permutations of n distinct objects when a particular object is not to be considered in the arrangement is given by n-1Pr.

⇒ The number of permutations of n distinct objects when a specific object is to be always included in the arrangement is given by r.n-1Pr-1.

⇒The number of combinations of n things taken r (0 < r < n) at a time is given by nCr = n!/r!(n - r)!

⇒ The relationship between combinations and permutations is nCr = nPr/r!

⇒ The number of ways of selecting r objects from n different objects subject to a certain condition like:

1. k particular objects are always included = n - kCr - k

2. k particular objects are never included = n - kCr

⇒ The number of arrangement of n distinct objects taken r at a time so that k particular objects are

(i) Always included = n - kCr - k.r!,

(ii) Never included = n - kCr.r!.

⇒ In order to compute the combination of n distinct items taken r at a time wherein, the chances of occurrence of any item are not fixed and maybe one, twice, thrice, …. up to r times is given by n + r - 1Cr

⇒ If there are m men and n women (m > n) and they have to be seated or accommodated in a row in such a way that no two women sit together then total no. of such arrangements = m + 1Cn m! This is also termed the Gap Method. 

⇒ If we have n different things taken r at a time in form of a garland or necklace, then the required number of arrangements is given by nCr(r - 1)!/2.

⇒ If there is a problem that requires n number of persons to be accommodated in such a way that a fixed number say ‘p’ are always together, then that particular set of p persons should be treated as one person. Hence, the total number of people in such a case becomes (n - m + 1). Therefore, the total number of possible arrangements is (n-m+1)! m! This is also termed the String Method.  

⇒ Let there be n types of objects with each type containing at least r objects. Then the number of ways of arranging r objects in a row is nr. 

⇒ The number of selections from n different objects, taking at least one = nC1 + nC2 + nC3 + ... + nCn = 2n - 1.  

⇒ The total number of selections of zero or more objects from n identical objects is n + 1.

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