Can someone help me with finding a solution for this ques...
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start from constraint given that w stands in front of y
method one - listing options
_y _ _ , 1*1*2*1=2
_ _ y _ , 2*1*1*1 OR 2*1*1*1 ==4
_ _ _ y , 3*2*1*1=6
total ways' count is (2+4+6)=12
method two - counting
fix y in the second place and count ways, 1 for w and 2C1 for other alphabets 1*2C1=2
fix y in the third place ~, 2 for w and 2C1 for other alphabets 2*2C1=4
fix y in fourth place ~, 3! ways for three alphabets=6
total ways's count is (2+4+6)=12
method one - listing options
_y _ _ , 1*1*2*1=2
_ _ y _ , 2*1*1*1 OR 2*1*1*1 ==4
_ _ _ y , 3*2*1*1=6
total ways' count is (2+4+6)=12
method two - counting
fix y in the second place and count ways, 1 for w and 2C1 for other alphabets 1*2C1=2
fix y in the third place ~, 2 for w and 2C1 for other alphabets 2*2C1=4
fix y in fourth place ~, 3! ways for three alphabets=6
total ways's count is (2+4+6)=12
geet_ge wrote:In how many ways can you arrange 4 alphabets(w,x,y & z) in a straight line so that w comes before y?
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Simple version...
Start as permutation: 4 * 3 * 2 * 1 = 24 ways to write the letters. Then, w will be in front of y exactly half of those times.
24/2 = 12 ways
Start as permutation: 4 * 3 * 2 * 1 = 24 ways to write the letters. Then, w will be in front of y exactly half of those times.
24/2 = 12 ways
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Hi
If W in the first position then total no ways=3!=6 ways
if W in the second position total ways=3p2 -1=5 ways
if w is in 3rd position only =1 way
total=6+5+1=12 ways.
If W in the first position then total no ways=3!=6 ways
if W in the second position total ways=3p2 -1=5 ways
if w is in 3rd position only =1 way
total=6+5+1=12 ways.