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junior college 2 | H2 Maths
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Date Posted: 5 years ago
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3ii)
To have at least one empty box between any two counters, there will be three gaps/spacings between the counters. Consider the following cases:
Case 1:
All gaps between counters are made of one empty box only.
1357, 2468, 3579
Case 2:
1 gap consists of two empty boxes, 2 gaps consist of one empty box.
1468,2579
1368,2479
1358,2469
Case 3:
2 gaps consist of two empty boxes , 1 gap consists of 1 empty box.
1479,1469,1369
Case 4:
1 gap consists of three empty boxes, 2 gaps consist of one empty box.
1579,1379,1359
number of ways = 15
Remember that two blue and two yellow counters are being used. Number of permutations = 4! ÷ 2! ÷ 2! = 6.
There are 6 permutations in each arrangement of the counters.
So total number of ways = 15 x 6 = 90.
To have at least one empty box between any two counters, there will be three gaps/spacings between the counters. Consider the following cases:
Case 1:
All gaps between counters are made of one empty box only.
1357, 2468, 3579
Case 2:
1 gap consists of two empty boxes, 2 gaps consist of one empty box.
1468,2579
1368,2479
1358,2469
Case 3:
2 gaps consist of two empty boxes , 1 gap consists of 1 empty box.
1479,1469,1369
Case 4:
1 gap consists of three empty boxes, 2 gaps consist of one empty box.
1579,1379,1359
number of ways = 15
Remember that two blue and two yellow counters are being used. Number of permutations = 4! ÷ 2! ÷ 2! = 6.
There are 6 permutations in each arrangement of the counters.
So total number of ways = 15 x 6 = 90.
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