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Question

Evaluate the following:
(i) 14C3
(ii) 12C10
(iii) 35C35
(iv) n + 1Cn
(v) r=15 5Cr.

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Solution

(i) We have,
14C3= 143 × 132× 121×11C0 [∵ nCr = nr n-1Cr-1]
14C3 = 364 [∵ nC0 = 1]

(ii) We have,
12C10=12C2 [∵ nCr = nCn-r]

12C10=12C2 = 122×111×10C0 [∵ nCr = nr n-1Cr-1]

12C10 = 122×111× 1 [∵ nC0 = 1]

12C10 = 66

(iii) We have,

35C35 = 35C0 [∵ nCr = nCn-r]

35C35 =1 [∵ nC0 = 1]


(iv) We have,

n + 1Cn =n + 1C1 [∵ nCr = nCn-r]
n+1Cn= n+1C1= n+11× nC0 [∵ nCr = nr n-1Cr-1]
n + 1Cn = n+1 [∵ nC0 = 1]

(v) We have,

r=15 5Cr=5C1+5C2+5C3+5C4+5C5

r=15 5Cr=5C1+5C3+5C3+5C1+5C0 [∵ nCr = nCn-r]

r=155Cr = 2×51×4C0 +2×53×42×31×2C0 + 5C0 [∵ nCr = nr n-1Cr-1]

r=155Cr = 10 + 20 + 1 = 31. [∵ nC0 = 1]

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