From 4 officers and 8 jawans, In how many ways can 6 be chosen so as to essentially include at least one officer ?

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Kumar Upendra Akshay

Math Educator for Std.11th ,12th , Engineering Entrance and Degree Level with 11+ Years Experience

total number of ways for selecting 6 persons out of 8 officers and 4 jawans are (i) 1 Officer and 5 jawans (ii) 2 Officers and 4 jawans (iii) 3 Officers and 3 jawans (iv) 4 Officers and 2 jawans 4C1. 8C5 + 4C2.8C4 + 4C3.8C3 + 4C4.8C2 = 224 + 420 + 224 + 28 = 896
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The required cases of at least one officer = total cases - cases having only jawans = 12C6 - 8C6 = 924 - 28 = 896.
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12C6 - 8C6
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12C6 - 8C6.. 1st I have selected 6 people out of 12 people (4 officers and 8 jawans) in 12C6 ways. We can select all 6 Jawans in 8C6 ways. However, we want atleast one officer. So it's 12C6-8C6 ways.
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Tutor - Maths and computer science

Total no of officers =4 Total no of jawan =8 Total no of committees =4C1×8C5 4C1=4!1!3!=4 8C5=8!5!3!=8×7×6×5!5!×3×2=56 ?4×56 ?224
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Solution: (4C1 x 8C5) + (4C2 x 8C4)+(4C3 x 8C3)+(4C4 x 8C2) ways
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The selection of 6 persons having at least one officer from a group of 4 officers and 8 jawans, can be possible in the following ways: 1) 1 officer + 5 jawans 2) 2 officer + 4 jawans 3) 3 officer + 3 jawans 4) 4 officer + 2 jawans Total number of ways = 4C1 x 8C5 + 4C2 x 8C4 + 4C3 x 8C3 + 4C4 x...
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The selection of 6 persons having at least one officer from a group of 4 officers and 8 jawans, can be possible in the following ways: 1) 1 officer + 5 jawans 2) 2 officer + 4 jawans 3) 3 officer + 3 jawans 4) 4 officer + 2 jawans Total number of ways = 4C1 x 8C5 + 4C2 x 8C4 + 4C3 x 8C3 + 4C4 x 8C2 = 896 read less
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required cases of at least one officer = total cases - cases having only jawans = 12C6 - 8C6 = 924 - 28 = 896
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896
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Total number of combinations - combinations in which there is no officer = 12c6 - 8c6 = 896
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