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NCERT Exemplar · Q42

Q.15C8+15C9−15C6−15C7={}^{15}C_{8} + {}^{15}C_{9} - {}^{15}C_{6} - {}^{15}C_{7} = ______.

Uttarakhand UbseShort· 2mImportance★★★★★est
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Use the symmetry property nCr=nCn−r{}^nC_r = {}^nC_{n-r} to rewrite terms, then group and apply Pascal's identity twice; the expression telescopes to 00.

The symmetry property of combinations tells us that choosing rr objects from nn is the same as choosing which n−rn-r objects to leave behind: nCr=nCn−r{}^nC_r = {}^nC_{n-r}. This seemingly simple fact is the key to unlocking this problem. Once we rewrite the higher-index terms using symmetry, the expression will reveal a hidden structure.

Let me apply the symmetry property 15Cr=15C15−r{}^{15}C_r = {}^{15}C_{15-r} to the terms with larger indices:

15C8=15C15−8=15C7{}^{15}C_8 = {}^{15}C_{15-8} = {}^{15}C_7

15C9=15C15−9=15C6{}^{15}C_9 = {}^{15}C_{15-9} = {}^{15}C_6

Now I can rewrite the original expression by substituting these equivalent forms:

15C8+15C9−15C6−15C7=15C7+15C6−15C6−15C7{}^{15}C_8 + {}^{15}C_9 - {}^{15}C_6 - {}^{15}C_7 = {}^{15}C_7 + {}^{15}C_6 - {}^{15}C_6 - {}^{15}C_7

Rearranging the terms to group like combinations:

=(15C7−15C7)+(15C6−15C6)= ({}^{15}C_7 - {}^{15}C_7) + ({}^{15}C_6 - {}^{15}C_6)

=0+0=0= 0 + 0 = 0 …

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