The individual women are represented by the letters A-H.
45 ways of seating 8 women on 10 seats (X is an empty seat):
ABCDEFGHXX ABCDEFGXHX ABCDEFXGHX ABCDEXFGHX ABCDXEFGHX
ABCXDEFGHX ABXCDEFGHX AXBCDEFGHX XABCDEFGHX ABCDEFGXXH
ABCDEFXGXH ABCDEXFGXH ABCDXEFGXH ABCXDEFGXH ABXCDEFGXH
AXBCDEFGXH XABCDEFGXH ABCDEFXXGH ABCDEXFXGH ABCDXEFXGH
ABCXDEFXGH ABXCDEFXGH AXBCDEFXGH XABCDEFXGH ABCDEXXFGH
ABCDXEXFGH ABCXDEXFGH ABXCDEXFGH AXBCDEXFGH XABCDEXFGH
ABCDXXEFGH ABCXDXEFGH ABXCDXEFGH AXBCDXEFGH XABCDXEFGH
ABCXXDEFGH ABXCXDEFGH AXBCXDEFGH XABCXDEFGH ABXXCDEFGH
AXBXCDEFGH XABXCDEFGH AXXBCDEFGH XAXBCDEFGH XXABCDEFGH
For each of these A-H, representing the individual women, can be arranged 40320 (8*7*6*5*4*3*2*1=40320) different ways. That would give us 45*40320 ways. But three women have to be seated together. Let's call these women A, B and C. Out of the 45 ways the women can be seated on 10 seats, how many have A, B and C together? There are 28 (shown in bold print). These three women can be arranged in 6 different ways:
ABC ACB BAC BCA CAB CBA.
The remaining 5 women can be in any arrangement, and there are 120 ways of arranging 5 women. If the three women have been specially identified out of the 8 women to sit together, then the total number of seating arrangements will be 6*120*28=20160.