There are five positions to be filled in a sports committee.

Question

There are five positions to be filled in a sports committee. For each position there are 3 different candidates recommended, one from each of the three states A,B and C. In how many ways can the five positions be filled such that there are more number of candidates from State B than from any other state?
(Each candidate can be recommended for only one position).

A. 40
B. 50
C. 51
D. 60
E. 61

Kushchokhani · Accepted Answer

Majority should be from B- B can be 3 or 4 or 5
Note that order is not important here. Just selection is reqd.

Case 1: BBBAA= \frac{5!}{3!2!} =10

Case 2: BBBCC= \frac{5!}{3!2!} =10

Case 3: BBBAC= \frac{5!}{3!} =20

Case 4: BBBBA= \frac{5!}{4!} =5

Case 5: BBBBC= \frac{5!}{4!} =5

Case 6: BBBBB= \frac{5!}{5!} =1

Adding above 6 cases, total cases = 51. Hence, answer is C.

Aabhash777 · Answer

Thank you so much. You have literally helped me alot. Did you take your practice from Jamboree as well? I have got still a few questions that I could not get correct

.

nahid78 · Answer

B can take either 3, 4 or 5 positions

if 3 then 5c3*2*2 (Forth and fifth position each can be selected in 2 ways)
if 4 the 5c4*2 (Fifth position is left for either A or C)
if 5 then 5c5
Total 40+10+1
=51

Iwillget770 · Answer

Why are you dividing by 3! and 2! in every case.

The question says that all the candidates are different for every position.

Say for CASE 1: Candidate B for 1st position is different from that of position 2 and 3. 
Similarly for A and C

Regards

KarishmaB : - Need your guidance

Bunuel · Answer

| (A, B, C) | (A, B, C) | (A, B, C) | (A, B, C) | (A, B, C) |

We have 5 positions to be filled, and for each position, there are candidates from the three states A, B, and C as shown above.

If we consider one possible case where four candidates from state B and one candidate from state A take the positions, then this can happen in the following 5 ways:

BBBBA
BBBAB
BBABB
BABBB
ABBBB

Hence, we should consider all the permutations of the 5 letters in "BBBBA", which is 5!/4!, to account for all the ways for this particular case. Similar logic applies to other cases.

Hope it's clear.

KarishmaB · Answer

As Bunuel has shown above, all places are distinct and we are arranging. Hence we are essentially multiplying by 5. (by using 5!/4!)
B for the first 4 positions and A for the last position. Or B for the first 3 and last position and A for the second last position etc.

This is similar to 5 coin tosses (each toss is different - 1st toss, 2nd toss etc) and in how many ways can we get more heads than tails? 
We can say HHHTT is one such case with probability 1/32. But another such case is HHTTH and another is HTTHH etc. So we multiply by 5!/3!*2!. We are essentially multiplying 1/32 by 10 to get all such cases. We are counting all possible arrangements.

Gemmie · Answer

1) Exactly 3 positions are filled by candidates from B
5C3 = 10
4th position: 2 ways to choose (candidate either from A or C)
5th position: 2 ways to choose (candidate either from A or C)

=> 10 * 2 * 2 = 40

2) Exactly 4 positions are filled by candidates from B
5C4 = 5
5th position: 2 ways to choose (candidate either from A or C)

=> 5 * 2 = 10

3) All 5 positions are candidates from B => 1 possibility

==> Total: 40 + 10 + 1 = 51

Jxmes · Answer

your calc suggests that order is important - which it is because the question implies that the positions on the committee are different. just adding this here in case someone's confused by the wording in that part of your response. unless i am wrong. if so, i'd appreciate if someone would correct me

There are five positions to be filled in a sports committee.

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There are five positions to be filled in a sports committee.

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